Question

a 0.38 m solution of ammonia was titrated with hydrochloric acid to the equivalence point, where the total volume was 1.50 times the original volume. at what ph does the equivalence point occur?
k_b = 1.8×10⁻⁵
ph =
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Explanation:

Step1: Find concentration of $NH_4^+$ at equivalence point

Let original volume be $V$, original concentration of $NH_3$ is $0.38\ M$. At equivalence, moles of $NH_3$ = moles of $HCl$, and final volume is $1.50V$. Moles of $NH_4^+$ formed = moles of $NH_3$ initially = $0.38V$. Concentration of $NH_4^+$ ($c$) = $\frac{0.38V}{1.50V} = \frac{0.38}{1.50} \approx 0.2533\ M$.

Step2: Calculate $K_a$ from $K_b$

We know $K_w = K_a \times K_b$, so $K_a = \frac{K_w}{K_b} = \frac{1.0\times10^{-14}}{1.8\times10^{-5}} \approx 5.56\times10^{-10}$.

Step3: Set up ICE table for $NH_4^+$ hydrolysis

$NH_4^+
ightleftharpoons NH_3 + H^+$
Initial: $0.2533\ M$ $0$ $0$
Change: $-x$ $+x$ $+x$
Equilibrium: $0.2533 - x$ $x$ $x$

Since $K_a$ is small, $0.2533 - x \approx 0.2533$. So $K_a = \frac{x^2}{0.2533}$, $x = \sqrt{K_a \times 0.2533} = \sqrt{5.56\times10^{-10} \times 0.2533} \approx \sqrt{1.408\times10^{-10}} \approx 1.187\times10^{-5}\ M$.

Step4: Calculate pH

$pH = -\log[H^+] = -\log(1.187\times10^{-5}) \approx 4.93$.

Answer:

$\approx 4.93$