Question

a 24.2 ml sample of 0.275 m ammonia, nh3, is titrated with 0.293 m nitric acid. after adding 32.9 ml of nitric acid, the ph is
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Explanation:

Step1: Calculate moles of $NH_3$ and $HNO_3$

Moles of $NH_3$: $n(NH_3) = C \times V = 0.275\ M \times 0.0242\ L = 0.006655\ mol$
Moles of $HNO_3$: $n(HNO_3) = 0.293\ M \times 0.0329\ L \approx 0.00964\ mol$

Step2: Determine limiting and excess reactants

The reaction is $NH_3 + HNO_3
ightarrow NH_4^+ + NO_3^-$.
Moles of $HNO_3$ exceed moles of $NH_3$. Excess $n(HNO_3) = 0.00964 - 0.006655 = 0.002985\ mol$

Step3: Calculate total volume

Total volume $V_{total} = 24.2\ mL + 32.9\ mL = 57.1\ mL = 0.0571\ L$

Step4: Calculate $[H^+]$ from excess $HNO_3$

$HNO_3$ is a strong acid, so $[H^+] = \frac{n_{excess}(HNO_3)}{V_{total}} = \frac{0.002985\ mol}{0.0571\ L} \approx 0.0523\ M$

Step5: Calculate pH

$pH = -\log[H^+] = -\log(0.0523) \approx 1.28$

Answer:

$\boxed{1.28}$