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AP Chemistry Flashcards: Strong Acids And Bases Ph Poh

Study Strong Acids And Bases Ph Poh 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

Strong Acids And Bases Ph Poh

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QUESTION
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State the formula for pH in terms of hydronium concentration.

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ANSWER

pH=−log⁡[H3O+]\mathrm{pH}=-\log[\mathrm{H_3O^+}]. Negative log converts small [H3O+][\mathrm{H_3O^+}] values to manageable pH scale.

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Flashcard 1: State the formula for pH in terms of hydronium concentration.

Answer: pH=−log⁡[H3O+]\mathrm{pH}=-\log[\mathrm{H_3O^+}]. Negative log converts small [H3O+][\mathrm{H_3O^+}] values to manageable pH scale.

Flashcard 2: What is the pH\text{pH} of a 0.010 M0.010\,\text{M} strong monoprotic acid?

Answer: pH=2.00\text{pH}=2.00. Monoprotic: n=1, so [H⁺]=0.010 M; pH=-log(0.010)=2.00.

Flashcard 3: What is the [OH−][\text{OH}^-] in a solution with [H+]=1.0×10−4 M[\text{H}^+]=1.0\times10^{-4}\,\text{M} at 25∘C25^\circ\text{C}?

Answer: [OH−]=1.0×10−10 M[\text{OH}^-]=1.0\times10^{-10}\,\text{M}. Use Kw=[H⁺][OH⁻]=1.0×10⁻¹⁴; [OH⁻]=1.0×10⁻¹⁴/1.0×10⁻⁴.

Flashcard 4: What is [H3O+][\mathrm{H_3O^+}] for a 0.030 M0.030\ \mathrm{M} solution of HCl\mathrm{HCl}?

Answer: [H3O+]=0.030 M[\mathrm{H_3O^+}]=0.030\ \mathrm{M}. HCl is monoprotic strong acid, so [H3O+][\mathrm{H_3O^+}] equals molarity.

Flashcard 5: What is the [H+][\text{H}^+] when pH=5.00\text{pH}=5.00?

Answer: [H+]=1.0×10−5 M[\text{H}^+]=1.0\times10^{-5}\,\text{M}. Apply [H⁺]=10⁻ᵖᴴ=10⁻⁵=1.0×10⁻⁵ M.

Flashcard 6: What is the formula for converting pH\text{pH} to [H+][\text{H}^+]?

Answer: [H+]=10−pH[\text{H}^+]=10^{-\text{pH}}. Inverse log operation converts pH back to concentration.

Flashcard 7: What is the formula that relates pOH\text{pOH} to [OH−][\text{OH}^-]?

Answer: pOH=−log⁡[OH−]\text{pOH}=-\log[\text{OH}^-]. Negative log converts hydroxide ion concentration to pOH scale.

Flashcard 8: What assumption is used for a strong base to relate [OH−][\text{OH}^-] to molarity?

Answer: [OH−]≈n Cbase[\text{OH}^-]\approx n\,C_{\text{base}}. Strong bases fully dissociate; n is OH⁻ ions per base molecule.

Flashcard 9: What is the pOH of a solution with [H3O+]=1.0×10−9 M[\mathrm{H_3O^+}]=1.0\times10^{-9}\ \mathrm{M} at 25∘C25^\circ\mathrm{C}?

Answer: pOH=5.00\mathrm{pOH}=5.00. pH=−log⁡(1.0×10−9)=9\mathrm{pH}=-\log(1.0\times10^{-9})=9; pOH=14−9=5\mathrm{pOH}=14-9=5.

Flashcard 10: What does nn represent in [H+]≈n Cacid[\text{H}^+]\approx n\,C_{\text{acid}} for a strong acid?

Answer: n=moles of H+ produced per mole of acidn=\text{moles of }\text{H}^+\text{ produced per mole of acid}. Counts H⁺ ions released per acid molecule (e.g., 2 for H₂SO₄).

Flashcard 11: What is the pH\text{pH} of a 0.10 M0.10\,\text{M} strong base that provides 3 OH−3\,\text{OH}^- per formula unit?

Answer: pH=13.48\text{pH}=13.48. n=3, so [OH⁻]=3×0.10=0.30 M; pOH=0.52; pH=14-0.52=13.48.

Flashcard 12: State the formula for pOH in terms of hydroxide concentration.

Answer: pOH=−log⁡[OH−]\mathrm{pOH}=-\log[\mathrm{OH^-}]. Negative log converts small [OH−][\mathrm{OH^-}] values to manageable pOH scale.

Flashcard 13: What assumption is used for a strong acid to relate [H+][\text{H}^+] to molarity?

Answer: [H+]≈n Cacid[\text{H}^+]\approx n\,C_{\text{acid}}. Strong acids fully dissociate; n is H⁺ ions per acid molecule.

Flashcard 14: What is the relationship between pH and pOH at 25∘C25^\circ\mathrm{C}?

Answer: pH+pOH=14.00\mathrm{pH}+\mathrm{pOH}=14.00. At 25°C25°\mathrm{C}, this sum equals −log⁡(Kw)=14.00-\log(K_w)=14.00.

Flashcard 15: What is the pOH of a 0.015 M0.015\ \mathrm{M} solution of Ba(OH)2\mathrm{Ba(OH)_2}?

Answer: pOH=1.52\mathrm{pOH}=1.52. pOH=−log⁡(0.030)=−log⁡(3×10−2)=1.52\mathrm{pOH}=-\log(0.030)=-\log(3\times10^{-2})=1.52.

Flashcard 16: What is the pH of a 1.0×10−3 M1.0\times10^{-3}\ \mathrm{M} solution of HNO3\mathrm{HNO_3}?

Answer: pH=3.00\mathrm{pH}=3.00. pH=−log⁡(1.0×10−3)=3.00\mathrm{pH}=-\log(1.0\times10^{-3})=3.00 for strong acid HNO₃.

Flashcard 17: What is the pH of a 2.0×10−5 M2.0\times10^{-5}\ \mathrm{M} solution of HBr\mathrm{HBr}?

Answer: pH=4.70\mathrm{pH}=4.70. pH=−log⁡(2.0×10−5)=5−log⁡(2)=4.70\mathrm{pH}=-\log(2.0\times10^{-5})=5-\log(2)=4.70.

Flashcard 18: What is the pOH\text{pOH} when pH=3.25\text{pH}=3.25 at 25∘C25^\circ\text{C}?

Answer: pOH=10.75\text{pOH}=10.75. Use pH+pOH=14.00; pOH=14.00-3.25=10.75.

Flashcard 19: Identify the pH of pure water at 25∘C25^\circ\mathrm{C}.

Answer: pH=7.00\mathrm{pH}=7.00. Neutral water has [H3O+]=[OH−]=1.0×10−7 M[\mathrm{H_3O^+}]=[\mathrm{OH^-}]=1.0\times10^{-7}\ \mathrm{M}.

Flashcard 20: What is [OH−][\mathrm{OH^-}] for a 0.015 M0.015\ \mathrm{M} solution of Ba(OH)2\mathrm{Ba(OH)_2}?

Answer: [OH−]=0.030 M[\mathrm{OH^-}]=0.030\ \mathrm{M}. Ba(OH)₂ releases 2 OH⁻ per unit: 0.015×2=0.030 M0.015\times^2=0.030\ \mathrm{M}.

Flashcard 21: What is the [OH−][\text{OH}^-] when pOH=9.00\text{pOH}=9.00?

Answer: [OH−]=1.0×10−9 M[\text{OH}^-]=1.0\times10^{-9}\,\text{M}. Apply [OH⁻]=10⁻ᵖᴼᴴ=10⁻⁹=1.0×10⁻⁹ M.

Flashcard 22: What is the pH of a 4.0×10−2 M4.0\times10^{-2}\ \mathrm{M} solution of NaOH\mathrm{NaOH} at 25∘C25^\circ\mathrm{C}?

Answer: pH=12.60\mathrm{pH}=12.60. pH=14.00−pOH=14.00−1.40=12.60\mathrm{pH}=14.00-\mathrm{pOH}=14.00-1.40=12.60.

Flashcard 23: What is the formula that relates pH\text{pH} to [H+][\text{H}^+]?

Answer: pH=−log⁡[H+]\text{pH}=-\log[\text{H}^+]. Negative log converts hydrogen ion concentration to pH scale.

Flashcard 24: Identify the assumption used for strong acids when finding [H3O+][\mathrm{H_3O^+}].

Answer: Strong acids dissociate ≈100%\approx100\%: [H3O+]≈[\mathrm{H_3O^+}]\approx acid molarity. Complete dissociation means initial acid concentration equals [H3O+][\mathrm{H_3O^+}].

Flashcard 25: Identify the assumption used for strong bases when finding [OH−][\mathrm{OH^-}].

Answer: Strong bases dissociate ≈100%\approx100\%: [OH−]≈[\mathrm{OH^-}]\approx base molarity ×\times stoichiometry. Complete dissociation; multiply by OH⁻ per formula unit.

Flashcard 26: What is KwK_w at 25∘C25^\circ\text{C} for water?

Answer: Kw=[H+][OH−]=1.0×10−14K_w=[\text{H}^+][\text{OH}^-]=1.0\times10^{-14}. Water's ion product constant at 25°C.

Flashcard 27: What is the formula for converting pOH\text{pOH} to [OH−][\text{OH}^-]?

Answer: [OH−]=10−pOH[\text{OH}^-]=10^{-\text{pOH}}. Inverse log operation converts pOH back to concentration.

Flashcard 28: What is the pH of a 3.0×10−4 M3.0\times10^{-4}\ \mathrm{M} solution of Ca(OH)2\mathrm{Ca(OH)_2} at 25∘C25^\circ\mathrm{C}?

Answer: pH=10.78\mathrm{pH}=10.78. [OH−]=2×3.0×10−4=6.0×10−4[\mathrm{OH^-}]=2\times^3.0\times10^{-4}=6.0\times10^{-4}; pOH=3.22\mathrm{pOH}=3.22; pH=10.78\mathrm{pH}=10.78.

Flashcard 29: What is the pOH\text{pOH} of a 0.0010 M0.0010\,\text{M} strong monohydroxide base?

Answer: pOH=3.00\text{pOH}=3.00. Monohydroxide: n=1, so [OH⁻]=0.0010 M; pOH=-log(0.0010)=3.00.

Flashcard 30: What is [H3O+][\mathrm{H_3O^+}] for a solution with pH=2.50\mathrm{pH}=2.50?

Answer: [H3O+]=10−2.50=3.2×10−3 M[\mathrm{H_3O^+}]=10^{-2.50}=3.2\times10^{-3}\ \mathrm{M}. Use [H3O+]=10−pH=10−2.50[\mathrm{H_3O^+}]=10^{-\mathrm{pH}}=10^{-2.50}.

Flashcard 31: What is the pH of a 0.015 M0.015\ \mathrm{M} solution of Ba(OH)2\mathrm{Ba(OH)_2} at 25∘C25^\circ\mathrm{C}?

Answer: pH=12.48\mathrm{pH}=12.48. pH=14.00−pOH=14.00−1.52=12.48\mathrm{pH}=14.00-\mathrm{pOH}=14.00-1.52=12.48.

Flashcard 32: What is the pH\text{pH} of a 0.0050 M0.0050\,\text{M} strong base that provides 2 OH−2\,\text{OH}^- per formula unit?

Answer: pH=12.00\text{pH}=12.00. n=2, so [OH⁻]=2×0.0050=0.010 M; pOH=2.00; pH=14-2=12.00.

Flashcard 33: What does nn represent in [OH−]≈n Cbase[\text{OH}^-]\approx n\,C_{\text{base}} for a strong base?

Answer: n=moles of OH− produced per mole of basen=\text{moles of }\text{OH}^-\text{ produced per mole of base}. Counts OH⁻ ions released per base molecule (e.g., 2 for Ba(OH)₂).

Flashcard 34: What is the pH\text{pH} when pOH=1.60\text{pOH}=1.60 at 25∘C25^\circ\text{C}?

Answer: pH=12.40\text{pH}=12.40. Use pH+pOH=14.00; pH=14.00-1.60=12.40.

Flashcard 35: What is the pOH of a 4.0×10−2 M4.0\times10^{-2}\ \mathrm{M} solution of NaOH\mathrm{NaOH}?

Answer: pOH=1.40\mathrm{pOH}=1.40. pOH=−log⁡(4.0×10−2)=2−log⁡(4)=1.40\mathrm{pOH}=-\log(4.0\times10^{-2})=2-\log(4)=1.40.

Flashcard 36: What is the pH\text{pH} of a 0.020 M0.020\,\text{M} strong diprotic acid?

Answer: pH=1.40\text{pH}=1.40. Diprotic: n=2, so [H⁺]=2×0.020=0.040 M; pH=-log(0.040)=1.40.

Flashcard 37: State the expression for KwK_w and its value at 25∘C25^\circ\mathrm{C}.

Answer: Kw=[H3O+][OH−]=1.0×10−14K_w=[\mathrm{H_3O^+}][\mathrm{OH^-}]=1.0\times10^{-14}. Water autoionization constant at 25°C25°\mathrm{C}.

Flashcard 38: What is [OH−][\mathrm{OH^-}] for a solution with pOH=5.20\mathrm{pOH}=5.20?

Answer: [OH−]=10−5.20=6.3×10−6 M[\mathrm{OH^-}]=10^{-5.20}=6.3\times10^{-6}\ \mathrm{M}. Use [OH−]=10−pOH=10−5.20[\mathrm{OH^-}]=10^{-\mathrm{pOH}}=10^{-5.20}.

Flashcard 39: What is the pH of a solution with [OH−]=1.0×10−12 M[\mathrm{OH^-}]=1.0\times10^{-12}\ \mathrm{M} at 25∘C25^\circ\mathrm{C}?

Answer: pH=2.00\mathrm{pH}=2.00. pOH=−log⁡(1.0×10−12)=12\mathrm{pOH}=-\log(1.0\times10^{-12})=12; pH=14−12=2\mathrm{pH}=14-12=2.

Flashcard 40: What is the relationship between pH\text{pH} and pOH\text{pOH} at 25∘C25^\circ\text{C}?

Answer: pH+pOH=14.00\text{pH}+\text{pOH}=14.00. At 25°C, the sum always equals 14 due to water's ion product.