Calculate the pH at the equivalence Calculate the pH at the equivalence point when 25.0 mL of 0.100 M NH3 is titrated by a 0.100 M HCl solution.point when 25.0 mL of 0.100 M NH3 is titrated by a 0.100 M HCl solution.Strategy The reaction between NH3 and HCl is NH3(aq) + HCl(aq) → NH4Cl(aq)We see that 1 mol NH3 1 mol HCl. At the equivalence point, the major species in solution are the salt NH4Cl

Question

Calculate the pH at the equivalence point when 25.0 mL of 0.100 M [latex]NH_{3}[/latex] is titrated by a 0.100 M HCl solution.

trategy The reaction between [latex]NH_{3}[/latex] and HCl is

[latex]NH_{3}(aq) + HCl(aq) → NH_{4}Cl(aq)[/latex]

We see that 1 mol [latex]NH_{3}[/latex] 1 mol HCl. At the equivalence point, the major species in solution are the salt [latex]NH_{4}Cl[/latex] (dissociated into [latex]NH^{+}_{4}[/latex] and [latex]Cl^{-}[/latex] ions) and [latex]H_{2}O[/latex]. First, we determine the concentration of [latex]NH_{4}Cl[/latex] formed. Then we calculate the pH as a result of the ion hydrolysis. The [latex]Cl^{-}[/latex] ion, being the conjugate base of a strong acid HCl, does not react with water. As usual, we ignore the ionization of water.

Accepted Answer

The number of moles of [latex]NH_{3}[/latex] in 25.0 mL of 0.100 M solution is

[latex]25.0 mL imes \frac{0.100 mol NH_{3}}{1 L NH_{3}} imes \frac{1 L}{1000 mL}=2.50 imes 10^{-3} mol [/latex]

At the equivalence point the number of moles of HCl added equals the number of moles of [latex]NH_{3}[/latex] . The changes in number of moles are summarized as follows:

[latex]\begin{matrix} & NH_{3}(aq) & + &HCl(aq) & ightarrow & NH_{4}Cl(aq)&& \ Initial (mol): & 2.50 imes 10^{-3} & &2.50 imes 10^{-3} &&0&&& \ Change (mol): & -2.50 imes 10^{-3} & &-2.50 imes 10^{-3} &&+2.50 imes 10^{-3} &&&& \ \hline Final (mol):&0&&0&&+2.50 imes 10^{-3} && \end{matrix}[/latex]

At the equivalence point, the concentrations of both the acid and the base are zero. The total volume is (25.0 + 25.0) mL, or 50.0 mL, so the concentration of the salt is

[latex][NH_{4}Cl]=\frac{2.50 imes 10^{-3} mol}{50.0 mL} imes \frac{1000 mL}{1 L} [/latex]

=0.0500mol/L=0.0500 M

The pH of the solution at the equivalence point is determined by the hydrolysis of [latex]NH^{+}_{4}[/latex] ions. We follow the procedure in Section 15.10.

Step 1: We represent the hydrolysis of the cation [latex]NH^{+}_{4}[/latex] , and let x be the equilibrium concentration of [latex]NH_{3}[/latex] and [latex]H^{+}[/latex] ions in mol/L:

[latex]\begin{matrix} & NH^{+}_{4}(aq) & \xrightleftharpoons[]{} &NH_{3}(aq) & +& H^{+}(aq)&& \ Initial (M): & 0.0500 & &0.000 &&0.000&&& \ Change (M): & -x& &+x&&+x&&&& \ \hline Equilibrium (M):&(0.0500-x)&&x&&x&& \end{matrix}[/latex]

Step 2: From Table 15.4 we obtain the [latex]K_{a}[/latex] for [latex]NH^{+}_{4}[/latex]

[latex]K_{a}=\frac{[NH_{3}][H^{+}]}{[NH^{+}_{4}]} [/latex]

[latex]5.6 imes 10^{-10}=\frac{x^{2} }{0.0500-x} [/latex]

Applying the approximation 0.0500 − x ≈ 0.0500, we get

[latex]5.6 imes 10^{-10}=\frac{x^{2} }{0.0500-x}\approx \frac{x^{2} }{0.0500}[/latex]

[latex]5.3 imes 10^{-6} M[/latex]

Thus, the pH is given by

[latex]pH=-log (5.3 imes 10^{-6} )[/latex]

=5.28

Check Note that the pH of the solution is acidic. This is what we would expect from the hydrolysis of the ammonium ion.

Question 16.6: Calculate the pH at the equivalence Calculate the pH at the ...

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