HNO₂ is a weak acid; it will not fully dissociate, so we need to use the HA → H⁺ + A^– reaction, with \(\displaystyle \dpi{100} \small K_{a}=\frac{\left [ products \right ]}{\left [ reactants \right ]}=\left \frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]}=4.1\times 10^{-4}\).

47.0g HNO₂ is equal to 1mol. 1mol into 4L gives a concentration of 0.25M when the acid is first dissolved; however, we want the pH at equilibrium, not at the initial state. As the acid dissolves, we know [HNO₂] will decrease to become ions, but we don't know by how much so we indicate the decrease as "x". As HNO₂ dissolves by a factor of x, the ion concentrations will increase by x.

                      HNO₂ → H⁺ + NO₂^– 

Initial             0.25M       0      0

Equilibrium     0.25 – x     x      x

Now, we can fill in our equation: \(\displaystyle \dpi{100} \small K_{a}=\left \frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]}=\frac{\left ( x \right )\left ( x \right )}{0.25-x}\).

Since x is very small, we can ignore it in the denominator: \(\displaystyle \dpi{100} \small K_{a}=\left \frac{\left ( x \right )\left ( x \right )}{\left (0.25 \right )}=4.1\times 10^{-4}\)

(they expect you to do this on the MCAT; you will never have to solve with x in the denominator on the exam!)

Solve for x, and you find \(\displaystyle x=1\times10^{-2}\). Looking at our table, we know that \(\displaystyle \dpi{100} \small x=\left [ H^{+} \right ]\)

Now we can solve for pH: \(\displaystyle pH=-log[H^+]=-log(1\times10^{-2})=2.0\)

Given the hydroxide ion concentration, we will need to work using pOH to find the pH. We know that the sum of pH and pOH is equal to 14.

\(\displaystyle pH + pOH = 14\)

\(\displaystyle pOH=-\log[OH^-]\)

Use our value for the concentration to find the pOH.

\(\displaystyle pOH=-\log(10^{-3})=3\)

Now that we have the pOH, we can use it to solve for the pH.

\(\displaystyle pH + 3 = 14\)

\(\displaystyle pH = 11\)

Since HCN is a weak acid, we must use the equilibrium equation.

\(\displaystyle K_{a}= \frac{[CN^{-}][H^{+}]}{[HCN]}\)

Because the HCN dissociates in solution, we expect the concentrations of protons and cyanide ions to increase, while the concentration of HCN will decrease. After determining the molarity of the solution, we can set up the equation below, using X as the amount of moles that dissociate.

\(\displaystyle \small K_{a}= \frac{[X][X]}{[[\frac{0.4mol}{2L}]-X]} = 6.2*10^{-10}\)

Because X is small, we can neglect its impact in the denominator.

\(\displaystyle \small K_{a}= \frac{[X][X]}{[\frac{0.4mol}{2L}]} = \frac{[X]^2}{[0.2]}=6.2*10^{-10}\)

\(\displaystyle \small X = 1.1*10^{-5}\)

Since X is the concentration of protons in the solution, we can calculate the pH by using the equation \(\displaystyle pH = -log[H^{+}]\).

\(\displaystyle pH=-log[1.1*10^{-5}]=4.95\)

The MCAT will typically allow you to approximate the difference in pH between two solutions. Look at the problem this way: you should already know a proton concentration of 1 * 10^-3M means a pH of 3. You do not need to know the exact pH of 5 * 10^-6M, but you should recognize that it is in between 1 * 10^-6M and 1 * 10^-5M. This means it will have a pH between 5 and 6.

The difference in pH will therefore be between 5 minus 3 and 6 minus 3.

\(\displaystyle pH\ difference=pH_A-pH_B\)

\(\displaystyle 5< pH_A< 6\)

\(\displaystyle 5-3< pH\ difference< 6-3\)

\(\displaystyle 2< pH\ difference< 3\)

As a result, look for the answer that is between 2 and 3. The only answer that makes sense is 2.3.

The equation to calculate pH is:

\(\displaystyle \small pH = - log\)\(\displaystyle \small \left [ H^{+}\right ]\)

The normal pH of arterial blood is around 7.4. This reflects a concentration of hydrogen ions that can be found using the pH equation.

\(\displaystyle \small 7.4 = - log\left [ H^{+}\right ]\)

\(\displaystyle \small 3.98 *10^{-8}M = \left [ H^{+}\right ]\)

Using similar calculations for the second blood sample, we can find the hydrogen ion concentration again.

\(\displaystyle 7.15=-log[H^+]\)

\(\displaystyle 7.08*10^{-8}M=[H^+]\)

Now that we have both concentrations, can find the ratio of the acidity of the two samples.

\(\displaystyle \frac{[H^+_2]}{[H^+_1]}=\frac{7.08*10^{-8}}{3.98*10^{-8}}=1.78\)

You may know from biological sciences that this is approaching a lethal level of acidosis.

\(\displaystyle HNO_3\) is a strong acid, meaning it will completely dissociate in solution. As such, the concentration of the acid will be equal to the proton concentration. Thus, to find pH, you should just plug the molar concentration of the acid solution into the pH formula.

\(\displaystyle [HNO_3]=[H^+]\)

\(\displaystyle pH=-log[H^+]\)

\(\displaystyle pH=-log(10^{-5})=5\)

A buffer must contain either a weak base and its salt or a weak acid and its salt. A mixture of ammonia and ammonium chloride is an example of the first case, since ammonia, NH₃, is a weak base and ammonium chloride, NH₄Cl, contains its salt.

Though autoionization of water produces small amounts of H₃O⁺ and OH^-, each conjugate salts of H₂O, they exist in such small amounts as to make any buffering effects negligible.

First, we need to determine from both the information presented in the paragraph and the chemical equation what HCO₃^- is doing. The paragraph tells us “HCO₃^- also [serves] to control the pH of the blood.” This is the definition of a buffer. A buffer is able to mitigate the addition of acid or base to a solution, or in this case, blood, by being deprotonated or protonated. For example, if the blood becomes too acidic, H⁺ will recombine with HCO₃^- to form H₂CO₃, thus stabilizing the pH. Alternatively, if the pH becomes too basic, H₂CO₃ will dissociate, releasing H⁺ into solution.

Whenever a concentration of the conjugate base is initially present in the solution, we say that the solution has been buffered. NaCN will completely dissolve in solution, meaning that there will be 1M of CN^- ions in the solution initially (one mole of NaCN in 1L of soultion equals 1M). With this in mind, the new equilibrium equation for the acid's dissociation is as follows:

\(\displaystyle K_a=\frac{[H^+][A^-]}{[HA]}=\frac{[H^+][CN^-]}{[HCN]}\)

The initial concentrations of CN^- and HCN are given in the question as 1M and 2M, respectively. The initial concentration of H⁺ is zero. As HCN begine to dissociate, H⁺ and CN^- will each increase by X amount, while HCN will decrease by X amount.

\(\displaystyle \small 6.2*10^{-10}= \frac{[X][1+X]}{[2-X]}\)

Since 1M and 2M are much larger numbers than the value for X, the X values next to these numbers can be omitted from the equation.

\(\displaystyle \small 6.2*10^{-10} = \frac{[X][1]}{[2]}\)

\(\displaystyle \small X = 1.24*10^{-9}\)

By plugging this value into the equation \(\displaystyle \small pH = -log[H^{+}]\), we determine that the pH of the solution is 8.9. (Remember that X is equal to the amount of increase in H⁺ ions).

\(\displaystyle pH=-log[1.24*10^{-9}]=8.9\)

This may seem wrong, because HCN is an acid, and we are used to acids resulting in solutions with pH values less than 7; however, HCN is a weak acid, and a large concentration of its conjugate base can result in a basic solution.

In a buffered solution, when the concentrations of acid and conjugate base are equal, we know the \(\displaystyle pH=pK_a\).

This is derived from the Henderson-Hasselbalch equation: \(\displaystyle pH = pK_{a} + \textup{log}_{10}(\frac{A^{-}}{HA})\).

When concentrations are equal, the log of \(\displaystyle 1\) is \(\displaystyle 0\), and \(\displaystyle pH=pK_a\).

\(\displaystyle [A^-]=[HA]\)

\(\displaystyle pH = pK_{a} + \textup{log}_{10}(1)=pK_a+0\)

In our question, the pH is approximately one unit greater than the \(\displaystyle pK_a\).

\(\displaystyle pH=4.97, pK_a=3.95\)

Because of this, we know that the log of the two concentrations must be equal to one.

\(\displaystyle pH=pK_a+log_{10}(\frac{[A^-]}{[HA]})\)

\(\displaystyle pH-pK_a=1=log_{10}(\frac{[A^-]}{[HA]})\)

The log of \(\displaystyle 10\) is \(\displaystyle 1\), and therefore the conjugate base must be ten times greater than the acid; therefore the ratio of acid to base is approximately \(\displaystyle 1:10\).

\(\displaystyle 1=log_{10}(10)=log_{10}(\frac{[A^-]}{[HA]})\)

\(\displaystyle \frac{[A^-]}{[HA]}=10\rightarrow [A^-]=10[HA]\)

We can quickly narrow this question to two possible answers, since the buffered solution is more basic than the \(\displaystyle pK_a\), and thus we would expect there to be a greater concentration of base than acid in the solution.