Question

A chemist titrates 190.0 mL of a 0.8125 M ammonia (NH) solution with 0.3733 M HCl solution at 25 °C. Calculate the pH at equivalence. The pK, of

ammonia is 4.75.
Round your answer to 2 decimal places.
Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HCl solution added.
pH ?

Answer 1

Approximately
`4.92`.

Step-by-step explanation:

Initial volume of the solution:
`V = 190.0\; \rm mL = 0.1900\; \rm L`.

Initial quantity of
`\rm NH_3`:

`\begin{aligned} n({\rm NH_3}) &= c({\rm NH_3}) \cdot V({\rm NH_3}) \\ &= 0.3733\; \rm mol \cdot L^(-1) * 0.1900\; \rm L \\ &\approx 0.154375\; \rm mol\end{aligned}`.

Ammonia
`\rm NH_3` reacts with hydrochloric
`\rm HCl` acid at a one-to-one ratio:

`\rm NH_3 + HCl \to NH_4 Cl`.

Hence, approximately
`n({\rm HCl}) = 0.154375\; \rm mol` of
`\rm HCl\!` molecules would be required to exactly react with the
`\rm NH_3\!` in the original solution and hence reach the equivalence point of this titration.

Calculate the volume of that
`0.3733\; \rm mol \cdot L^(-1)`
`\rm HCl` solution required for reaching the equivalence point of this titration:

`\begin{aligned}V({\rm HCl}) &= \frac{n({\rm HCl})}{c({\rm HCl})} \\ &\approx (0.154375\; \rm mol)/(0.3733\; \rm mol \cdot L^(-1)) \approx 0.413541\; \rm L\end{aligned}`.

Hence, by the assumption stated in the question, the volume of the solution at the equivalence point would be approximately
`0.413541\; \rm L + 0.1900\; \rm L \approx 0.6035\; \rm L`.

If no hydrolysis took place,
`0.154375\; \rm mol` of
`\rm NH_4 Cl` would be produced. Because
`\rm NH_4 Cl\!` is a soluble salt, the solution would contain
`0.154375\; \rm mol\!` of
`\rm {NH_4}^(+)` ions. The concentration of
`\rm {NH_4}^(+)\!` would be approximately:

`\begin{aligned}c({\rm {NH_4}^(+)}) &= \frac{n({\rm {NH_4}^(+)})}{V({\rm {NH_4}^(+)})}\\ &\approx (0.154375\; \rm mol)/(0.6035\; \rm L) \approx 0.255782\; \rm mol \cdot L^(-1)\end{aligned}`.

However, because
`\rm NH_3 \cdot H_2O` is a weak base, its conjugate
`\rm {NH_4}^(+)` would be a weak base.

`\begin{aligned}pK_(\rm a)({{\rm NH_4}}^(+)) &= pK_(\rm w) - pK_(\rm b)({\rm NH_3})\\ &\approx 13.99 - 4.75 = 9.25\end{aligned}`.

Hence, the following reversible reaction would be take place in the solution at the equivalence point:

`\rm {NH_4}^(+) \rightleftharpoons NH_3 + H^(+)`.

Let
`x\; \rm mol \cdot L^(-1)` be the increase in the concentration of
`\rm H^(+)` in this solution because of this reversible reaction. (Notice that
`x \ge 0`.) Construct the following
`\text{RICE}` table:

`\begin{array}c \textbf{R}& \rm {\rm NH_4}^(+) & \rightleftharpoons & {\rm NH_3}& + & {\rm H}^(+)\\ \textbf{I} & 0.255782 \; \rm M \\ \textbf{C} & -x \;\rm M & & + x\;\rm M & & + x\; \rm M \\ \textbf{E} & (0.255782 - x)\; \rm M & & x\; \rm M & & x\; \rm M\end{array}`.

Thus, at equilibrium:

• Concentration of the weak acid:
  `[{\rm {NH_4}^(+)}] \approx (0.255782 - x) \; \rm M`.
• Concentration of the conjugate of the weak acid:
  `[{\rm NH_3}] = x\; \rm M`.
• Concentration of
  `\rm H^(+)`:
  `[{\rm {H}^(+)}] \approx x\; \rm M`.

`\displaystyle \frac{[{\rm NH_3}] \cdot [{\rm H^(+)}]}{[{ \rm {NH_4}^(+)}]} = 10^{pK_\text{a}({\rm {NH_4}^(+)})}`.

`\displaystyle (x^2)/(0.255782 - x) \approx 10^(-9.25)`

Solve for
`x`. (Notice that the value of
`x\!` is likely to be much smaller than
`0.255782`. Hence, the denominator on the left-hand side
`(0.255782 - x) \approx 0.255782`.)

`x \approx 1.19929 * 10^(-5)`.

Hence, the concentration of
`\rm H^(+)` at the equivalence point of this titration would be approximately
`1.19929 * 10^(-5)\; \rm M`.

Hence, the
`pH` at the equivalence point of this titration would be:

`\begin{aligned}pH &= -\log_(10)[{\rm {H}^(+)}] \\ &\approx -\log_(10) \left(1.19929 * 10^(-5)\right) \approx 4.92\end{aligned}`.