AP Chemistry Flashcards: Ph And Solubility

Study Ph And Solubility in AP Chemistry with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

AP Chemistry

Ph And Solubility

0 mastered0 still learning

0% Complete

QUESTION
1/ 146

Identify the pH of a neutral solution at 25°C.

Tap card or press Space to flip

ANSWER

pH = 7. Equal concentrations of [H+][H^+] and [OH][OH^-].

How well did you know it?

Card 1 / 146

What this deck covers

This deck focuses on Ph And Solubility, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

All flashcards

Flashcard 1: Identify the pH of a neutral solution at 25°C.

Answer: pH = 7. Equal concentrations of [H+][H^+] and [OH][OH^-].

Flashcard 2: What is the pH of a solution with [H+]=1×107 M[H^+] = 1 \times 10^{-7} \text{ M}?

Answer: pH = 7. pH=log(107)=7pH = -\log(10^{-7}) = 7

Flashcard 3: Determine the pOH of a solution with [OH]=1×104 M[OH^-] = 1 \times 10^{-4} \text{ M}.

Answer: pOH = 4. pOH=log(104)=4pOH = -\log(10^{-4}) = 4

Flashcard 4: What is the solubility product constant for PbI2PbI_2?

Answer: Ksp=[Pb2+][I]2K_{sp} = [Pb^{2+}][I^-]^2. Iodide ion squared due to formula stoichiometry.

Flashcard 5: What is the pH of a 0.1 M0.1 \text{ M} NaOHNaOH solution?

Answer: pH = 13. Strong base: [OH]=0.1[OH^-] = 0.1, pOH=1pOH = 1

Flashcard 6: Identify the pH of a neutral solution at 25°C.

Answer: pH = 7. Equal concentrations of [H+][H^+] and [OH][OH^-].

Flashcard 7: State the KspK_{sp} expression for ZnSZnS.

Answer: Ksp=[Zn2+][S2]K_{sp} = [Zn^{2+}][S^{2-}]. 1:1 stoichiometry for ion products.

Flashcard 8: What is the solubility product constant for PbI2PbI_2?

Answer: Ksp=[Pb2+][I]2K_{sp} = [Pb^{2+}][I^-]^2. Iodide ion squared due to formula stoichiometry.

Flashcard 9: What is the definition of the solubility product constant (KspK_{sp})?

Answer: KspK_{sp} is the equilibrium constant for a solid dissolving in water. Represents dissolution equilibrium of ionic solids.

Flashcard 10: What is the definition of the solubility product constant (KspK_{sp})?

Answer: KspK_{sp} is the equilibrium constant for a solid dissolving in water. Represents dissolution equilibrium of ionic solids.

Flashcard 11: What is the KspK_{sp} expression for CuSCuS?

Answer: Ksp=[Cu2+][S2]K_{sp} = [Cu^{2+}][S^{2-}]. 1:1 stoichiometry for ion products.

Flashcard 12: State the formula for calculating pOH from [OH][OH^-].

Answer: pOH=log[OH]pOH = -\text{log}[OH^-]. Take negative log of hydroxide ion concentration.

Flashcard 13: Determine the pH of a 0.025 M0.025 \text{ M} HBrHBr solution.

Answer: pH = 1.6. pH=log(0.025)=1.6pH = -\log(0.025) = 1.6

Flashcard 14: Which solution has a higher pH: 0.1 M0.1 \text{ M} HClHCl or 0.1 M0.1 \text{ M} NaOHNaOH?

Answer: 0.1 M0.1 \text{ M} NaOHNaOH. NaOHNaOH is basic with higher pH value.

Flashcard 15: Find the pH of a 0.075 M0.075 \text{ M} Ba(OH)2Ba(OH)_2 solution.

Answer: pH = 13.18. Diprotic base: [OH]=0.15[OH^-] = 0.15, pOH=0.82pOH = 0.82

Flashcard 16: Calculate the pH of pure water at 25°C.

Answer: pH = 7. [H+]=1.0×107[H^+] = 1.0 \times 10^{-7} at equilibrium.

Flashcard 17: What is the KspK_{sp} expression for CaF2CaF_2?

Answer: Ksp=[Ca2+][F]2K_{sp} = [Ca^{2+}][F^-]^2. Fluoride ion squared due to formula stoichiometry.

Flashcard 18: State the formula for converting pOH to pH.

Answer: pH=14pOHpH = 14 - pOH. Derived from pH+pOH=14pH + pOH = 14 relationship.

Flashcard 19: What is the KspK_{sp} expression for CuSCuS?

Answer: Ksp=[Cu2+][S2]K_{sp} = [Cu^{2+}][S^{2-}]. 1:1 stoichiometry for ion products.

Flashcard 20: What is the pH of a 0.02 M0.02 \text{ M} KOHKOH solution?

Answer: pH = 12.3. [OH]=0.02[OH^-] = 0.02, pOH=1.7pOH = 1.7

Flashcard 21: What is the KspK_{sp} expression for PbSO4PbSO_4?

Answer: Ksp=[Pb2+][SO42]K_{sp} = [Pb^{2+}][SO_4^{2-}]. 1:1 stoichiometry for ion products.

Flashcard 22: What is the KspK_{sp} expression for Mg(OH)2Mg(OH)_2?

Answer: Ksp=[Mg2+][OH]2K_{sp} = [Mg^{2+}][OH^-]^2. Hydroxide ion squared due to formula stoichiometry.

Flashcard 23: What is the pH of a solution with [H+]=1×103 M[H^+] = 1 \times 10^{-3} \text{ M}?

Answer: pH = 3. pH=log(103)=3pH = -\log(10^{-3}) = 3

Flashcard 24: State the ion product constant for water (KwK_w) at 25°C.

Answer: Kw=1.0×1014K_w = 1.0 \times 10^{-14}. Equilibrium constant for water autoionization.

Flashcard 25: Identify the KspK_{sp} expression for BaSO4BaSO_4.

Answer: Ksp=[Ba2+][SO42]K_{sp} = [Ba^{2+}][SO_4^{2-}]. 1:1 stoichiometry for ion products.

Flashcard 26: Identify the KspK_{sp} expression for BaSO4BaSO_4.

Answer: Ksp=[Ba2+][SO42]K_{sp} = [Ba^{2+}][SO_4^{2-}]. 1:1 stoichiometry for ion products.

Flashcard 27: What is the pH of a 0.02 M0.02 \text{ M} KOHKOH solution?

Answer: pH = 12.3. [OH]=0.02[OH^-] = 0.02, pOH=1.7pOH = 1.7

Flashcard 28: What is the KspK_{sp} expression for CaCO3CaCO_3?

Answer: Ksp=[Ca2+][CO32]K_{sp} = [Ca^{2+}][CO_3^{2-}]. 1:1 stoichiometry for ion products.

Flashcard 29: What is the KspK_{sp} expression for Ag2CrO4Ag_2CrO_4?

Answer: Ksp=[Ag+]2[CrO42]K_{sp} = [Ag^+]^2[CrO_4^{2-}]. Silver ion squared due to formula stoichiometry.

Flashcard 30: What is the relationship between pH and pOH at 25°C?

Answer: pH+pOH=14pH + pOH = 14. Sum equals 14 at standard temperature.

Flashcard 31: What is the definition of pH?

Answer: pH is the negative logarithm of the hydrogen ion concentration: pH=log[H+]pH = -\text{log}[H^+]. It measures acidity using logarithmic scale.

Flashcard 32: What is the pH of a 0.02 M0.02 \text{ M} KOHKOH solution?

Answer: pH = 12.3. [OH]=0.02[OH^-] = 0.02, pOH=1.7pOH = 1.7

Flashcard 33: State the formula for calculating pH from [H+][H^+].

Answer: pH=log[H+]pH = -\text{log}[H^+]. Take negative log of hydrogen ion concentration.

Flashcard 34: If [OH]=1×109 M[OH^-] = 1 \times 10^{-9} \text{ M}, what is the pH of the solution?

Answer: pH = 5. pOH=9pOH = 9, so pH=149=5pH = 14 - 9 = 5

Flashcard 35: What is the solubility product constant for PbI2PbI_2?

Answer: Ksp=[Pb2+][I]2K_{sp} = [Pb^{2+}][I^-]^2. Iodide ion squared due to formula stoichiometry.

Flashcard 36: What is the relationship between pH and pOH at 25°C?

Answer: pH+pOH=14pH + pOH = 14. Sum equals 14 at standard temperature.

Flashcard 37: What is the solubility product constant (KspK_{sp}) expression for AgClAgCl?

Answer: Ksp=[Ag+][Cl]K_{sp} = [Ag^+][Cl^-]. Products of ion concentrations at equilibrium.

Flashcard 38: What is the KspK_{sp} expression for Hg2I2Hg_2I_2?

Answer: Ksp=[Hg22+][I]2K_{sp} = [Hg_2^{2+}][I^-]^2. Iodide ion squared due to formula stoichiometry.

Flashcard 39: What is the solubility product constant for PbI2PbI_2?

Answer: Ksp=[Pb2+][I]2K_{sp} = [Pb^{2+}][I^-]^2. Iodide ion squared due to formula stoichiometry.

Flashcard 40: What is the KspK_{sp} expression for Mg(OH)2Mg(OH)_2?

Answer: Ksp=[Mg2+][OH]2K_{sp} = [Mg^{2+}][OH^-]^2. Hydroxide ion squared due to formula stoichiometry.

Flashcard 41: Calculate the pH of a 0.001 M0.001 \text{ M} HNO3HNO_3 solution.

Answer: pH = 3. Strong acid: pH=log(0.001)=3pH = -\log(0.001) = 3

Flashcard 42: Calculate the pH of a 0.1 M0.1 \text{ M} CH3COOHCH_3COOH solution, Ka=1.8×105K_a = 1.8 \times 10^{-5}.

Answer: pH ≈ 2.87. Weak acid requires ICE table calculation.

Flashcard 43: What is the definition of pH?

Answer: pH is the negative logarithm of the hydrogen ion concentration: pH=log[H+]pH = -\text{log}[H^+]. It measures acidity using logarithmic scale.

Flashcard 44: Determine the pOH of a solution with [OH]=1×104 M[OH^-] = 1 \times 10^{-4} \text{ M}.

Answer: pOH = 4. pOH=log(104)=4pOH = -\log(10^{-4}) = 4

Flashcard 45: What is the KspK_{sp} expression for CuSCuS?

Answer: Ksp=[Cu2+][S2]K_{sp} = [Cu^{2+}][S^{2-}]. 1:1 stoichiometry for ion products.

Flashcard 46: What is the relationship between pH and pOH at 25°C?

Answer: pH+pOH=14pH + pOH = 14. Sum equals 14 at standard temperature.

Flashcard 47: Find the pH of a 0.03 M0.03 \text{ M} LiOHLiOH solution.

Answer: pH = 12.5. [OH]=0.03[OH^-] = 0.03, pOH=1.52pOH = 1.52

Flashcard 48: State the ion product constant for water (KwK_w) at 25°C.

Answer: Kw=1.0×1014K_w = 1.0 \times 10^{-14}. Equilibrium constant for water autoionization.

Flashcard 49: Determine the pH of a 0.025 M0.025 \text{ M} HBrHBr solution.

Answer: pH = 1.6. pH=log(0.025)=1.6pH = -\log(0.025) = 1.6

Flashcard 50: What is the KspK_{sp} expression for Ag2CrO4Ag_2CrO_4?

Answer: Ksp=[Ag+]2[CrO42]K_{sp} = [Ag^+]^2[CrO_4^{2-}]. Silver ion squared due to formula stoichiometry.

Flashcard 51: Determine the pOH of a solution with [OH]=1×104 M[OH^-] = 1 \times 10^{-4} \text{ M}.

Answer: pOH = 4. pOH=log(104)=4pOH = -\log(10^{-4}) = 4

Flashcard 52: What is the KspK_{sp} expression for Ca(OH)2Ca(OH)_2?

Answer: Ksp=[Ca2+][OH]2K_{sp} = [Ca^{2+}][OH^-]^2. Hydroxide ion squared due to formula stoichiometry.

Flashcard 53: What is the KspK_{sp} expression for Hg2I2Hg_2I_2?

Answer: Ksp=[Hg22+][I]2K_{sp} = [Hg_2^{2+}][I^-]^2. Iodide ion squared due to formula stoichiometry.

Flashcard 54: Which solution has a higher pH: 0.1 M0.1 \text{ M} HClHCl or 0.1 M0.1 \text{ M} NaOHNaOH?

Answer: 0.1 M0.1 \text{ M} NaOHNaOH. NaOHNaOH is basic with higher pH value.

Flashcard 55: Determine the KspK_{sp} expression for Fe(OH)3Fe(OH)_3.

Answer: Ksp=[Fe3+][OH]3K_{sp} = [Fe^{3+}][OH^-]^3. Hydroxide ion cubed due to formula stoichiometry.

Flashcard 56: What is the KspK_{sp} expression for Ca(OH)2Ca(OH)_2?

Answer: Ksp=[Ca2+][OH]2K_{sp} = [Ca^{2+}][OH^-]^2. Hydroxide ion squared due to formula stoichiometry.

Flashcard 57: What is the pH of a 0.01 M0.01 \text{ M} HClHCl solution?

Answer: pH = 2. Strong acid completely ionizes: [H+]=0.01[H^+] = 0.01

Flashcard 58: Determine the pOH of a solution with [OH]=1×104 M[OH^-] = 1 \times 10^{-4} \text{ M}.

Answer: pOH = 4. pOH=log(104)=4pOH = -\log(10^{-4}) = 4

Flashcard 59: Determine the KspK_{sp} expression for Fe(OH)3Fe(OH)_3.

Answer: Ksp=[Fe3+][OH]3K_{sp} = [Fe^{3+}][OH^-]^3. Hydroxide ion cubed due to formula stoichiometry.

Flashcard 60: Identify the pH of a neutral solution at 25°C.

Answer: pH = 7. Equal concentrations of [H+][H^+] and [OH][OH^-].

Flashcard 61: What is the KspK_{sp} expression for Ag2SO4Ag_2SO_4?

Answer: Ksp=[Ag+]2[SO42]K_{sp} = [Ag^+]^2[SO_4^{2-}]. Silver ion squared due to formula stoichiometry.

Flashcard 62: Find the pH of a 0.03 M0.03 \text{ M} LiOHLiOH solution.

Answer: pH = 12.5. [OH]=0.03[OH^-] = 0.03, pOH=1.52pOH = 1.52

Flashcard 63: State the formula for calculating pOH from [OH][OH^-].

Answer: pOH=log[OH]pOH = -\text{log}[OH^-]. Take negative log of hydroxide ion concentration.

Flashcard 64: State the formula for converting pOH to pH.

Answer: pH=14pOHpH = 14 - pOH. Derived from pH+pOH=14pH + pOH = 14 relationship.

Flashcard 65: Calculate the pH of a 0.005 M0.005 \text{ M} HClHCl solution.

Answer: pH = 2.3. pH=log(0.005)=2.3pH = -\log(0.005) = 2.3

Flashcard 66: Which solution has a higher pH: 0.1 M0.1 \text{ M} HClHCl or 0.1 M0.1 \text{ M} NaOHNaOH?

Answer: 0.1 M0.1 \text{ M} NaOHNaOH. NaOHNaOH is basic with higher pH value.

Flashcard 67: What is the definition of pH?

Answer: pH is the negative logarithm of the hydrogen ion concentration: pH=log[H+]pH = -\text{log}[H^+]. It measures acidity using logarithmic scale.

Flashcard 68: What is the pH of a 0.1 M0.1 \text{ M} NaOHNaOH solution?

Answer: pH = 13. Strong base: [OH]=0.1[OH^-] = 0.1, pOH=1pOH = 1

Flashcard 69: What is the KspK_{sp} expression for Ca(OH)2Ca(OH)_2?

Answer: Ksp=[Ca2+][OH]2K_{sp} = [Ca^{2+}][OH^-]^2. Hydroxide ion squared due to formula stoichiometry.

Flashcard 70: What is the KspK_{sp} expression for Mg(OH)2Mg(OH)_2?

Answer: Ksp=[Mg2+][OH]2K_{sp} = [Mg^{2+}][OH^-]^2. Hydroxide ion squared due to formula stoichiometry.

Flashcard 71: Determine the KspK_{sp} expression for Fe(OH)3Fe(OH)_3.

Answer: Ksp=[Fe3+][OH]3K_{sp} = [Fe^{3+}][OH^-]^3. Hydroxide ion cubed due to formula stoichiometry.

Flashcard 72: Calculate the pH of pure water at 25°C.

Answer: pH = 7. [H+]=1.0×107[H^+] = 1.0 \times 10^{-7} at equilibrium.

Flashcard 73: What is the pH of a 0.02 M0.02 \text{ M} KOHKOH solution?

Answer: pH = 12.3. [OH]=0.02[OH^-] = 0.02, pOH=1.7pOH = 1.7

Flashcard 74: Calculate the pH of a 0.001 M0.001 \text{ M} HNO3HNO_3 solution.

Answer: pH = 3. Strong acid: pH=log(0.001)=3pH = -\log(0.001) = 3

Flashcard 75: Which is more soluble in water: AgClAgCl or NaClNaCl?

Answer: NaClNaCl. NaClNaCl is highly soluble ionic compound.

Flashcard 76: What is the KspK_{sp} expression for Ag2SO4Ag_2SO_4?

Answer: Ksp=[Ag+]2[SO42]K_{sp} = [Ag^+]^2[SO_4^{2-}]. Silver ion squared due to formula stoichiometry.

Flashcard 77: What is the KspK_{sp} expression for PbSO4PbSO_4?

Answer: Ksp=[Pb2+][SO42]K_{sp} = [Pb^{2+}][SO_4^{2-}]. 1:1 stoichiometry for ion products.

Flashcard 78: Find the pH of a 0.075 M0.075 \text{ M} Ba(OH)2Ba(OH)_2 solution.

Answer: pH = 13.18. Diprotic base: [OH]=0.15[OH^-] = 0.15, pOH=0.82pOH = 0.82

Flashcard 79: State the formula for calculating pOH from [OH][OH^-].

Answer: pOH=log[OH]pOH = -\text{log}[OH^-]. Take negative log of hydroxide ion concentration.

Flashcard 80: Which solution has a higher pH: 0.1 M0.1 \text{ M} HClHCl or 0.1 M0.1 \text{ M} NaOHNaOH?

Answer: 0.1 M0.1 \text{ M} NaOHNaOH. NaOHNaOH is basic with higher pH value.

Flashcard 81: State the ion product constant for water (KwK_w) at 25°C.

Answer: Kw=1.0×1014K_w = 1.0 \times 10^{-14}. Equilibrium constant for water autoionization.

Flashcard 82: Find the pH of a 0.03 M0.03 \text{ M} LiOHLiOH solution.

Answer: pH = 12.5. [OH]=0.03[OH^-] = 0.03, pOH=1.52pOH = 1.52

Flashcard 83: What is the definition of pH?

Answer: pH is the negative logarithm of the hydrogen ion concentration: pH=log[H+]pH = -\text{log}[H^+]. It measures acidity using logarithmic scale.

Flashcard 84: What is the pH of a 0.1 M0.1 \text{ M} NaOHNaOH solution?

Answer: pH = 13. Strong base: [OH]=0.1[OH^-] = 0.1, pOH=1pOH = 1

Flashcard 85: What is the pH of a solution with [H+]=1×103 M[H^+] = 1 \times 10^{-3} \text{ M}?

Answer: pH = 3. pH=log(103)=3pH = -\log(10^{-3}) = 3

Flashcard 86: Which is more soluble in water: AgClAgCl or NaClNaCl?

Answer: NaClNaCl. NaClNaCl is highly soluble ionic compound.

Flashcard 87: What is the KspK_{sp} expression for Ag2CrO4Ag_2CrO_4?

Answer: Ksp=[Ag+]2[CrO42]K_{sp} = [Ag^+]^2[CrO_4^{2-}]. Silver ion squared due to formula stoichiometry.

Flashcard 88: Calculate the pH of a 0.1 M0.1 \text{ M} CH3COOHCH_3COOH solution, Ka=1.8×105K_a = 1.8 \times 10^{-5}.

Answer: pH ≈ 2.87. Weak acid requires ICE table calculation.

Flashcard 89: What is the KspK_{sp} expression for CaF2CaF_2?

Answer: Ksp=[Ca2+][F]2K_{sp} = [Ca^{2+}][F^-]^2. Fluoride ion squared due to formula stoichiometry.

Flashcard 90: What is the solubility product constant (KspK_{sp}) expression for AgClAgCl?

Answer: Ksp=[Ag+][Cl]K_{sp} = [Ag^+][Cl^-]. Products of ion concentrations at equilibrium.

Flashcard 91: State the formula for calculating pH from [H+][H^+].

Answer: pH=log[H+]pH = -\text{log}[H^+]. Take negative log of hydrogen ion concentration.

Flashcard 92: What is the KspK_{sp} expression for Ca(OH)2Ca(OH)_2?

Answer: Ksp=[Ca2+][OH]2K_{sp} = [Ca^{2+}][OH^-]^2. Hydroxide ion squared due to formula stoichiometry.

Flashcard 93: Calculate the pH of a 0.001 M0.001 \text{ M} HNO3HNO_3 solution.

Answer: pH = 3. Strong acid: pH=log(0.001)=3pH = -\log(0.001) = 3

Flashcard 94: What is the definition of the solubility product constant (KspK_{sp})?

Answer: KspK_{sp} is the equilibrium constant for a solid dissolving in water. Represents dissolution equilibrium of ionic solids.

Flashcard 95: What is the solubility product constant (KspK_{sp}) expression for AgClAgCl?

Answer: Ksp=[Ag+][Cl]K_{sp} = [Ag^+][Cl^-]. Products of ion concentrations at equilibrium.

Flashcard 96: State the formula for calculating pOH from [OH][OH^-].

Answer: pOH=log[OH]pOH = -\text{log}[OH^-]. Take negative log of hydroxide ion concentration.

Flashcard 97: What is the KspK_{sp} expression for CaF2CaF_2?

Answer: Ksp=[Ca2+][F]2K_{sp} = [Ca^{2+}][F^-]^2. Fluoride ion squared due to formula stoichiometry.

Flashcard 98: Determine the KspK_{sp} expression for Fe(OH)3Fe(OH)_3.

Answer: Ksp=[Fe3+][OH]3K_{sp} = [Fe^{3+}][OH^-]^3. Hydroxide ion cubed due to formula stoichiometry.

Flashcard 99: Identify the KspK_{sp} expression for BaSO4BaSO_4.

Answer: Ksp=[Ba2+][SO42]K_{sp} = [Ba^{2+}][SO_4^{2-}]. 1:1 stoichiometry for ion products.

Flashcard 100: Calculate the pH of pure water at 25°C.

Answer: pH = 7. [H+]=1.0×107[H^+] = 1.0 \times 10^{-7} at equilibrium.