QuestionJuly 15, 2025

30.0g of N_(2)CSH_(2)Cl is mixed with a 1.00L.solution of 0.100MN_(2)CSH. If 0.200 mol of NaOH is added, what will be the final pH? N_(2)CSHKb=6.1^ast 10^-6

30.0g of N_(2)CSH_(2)Cl is mixed with a 1.00L.solution of 0.100MN_(2)CSH. If 0.200 mol of NaOH is added, what will be the final pH? N_(2)CSHKb=6.1^ast 10^-6
30.0g of N_(2)CSH_(2)Cl is mixed with a 1.00L.solution of 0.100MN_(2)CSH. If 0.200 mol of NaOH is added, what will be the final pH? N_(2)CSHKb=6.1^ast 10^-6

Solution
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Answer

8.63 Explanation 1. Calculate moles of N_{2}CSH_{2}Cl Molar mass of N_{2}CSH_{2}Cl is approximately 95.54 g/mol. Moles = \frac{30.0 \text{ g}}{95.54 \text{ g/mol}} = 0.314 \text{ mol}. 2. Determine initial concentrations Initial concentration of N_{2}CSH_{2}^+ = \frac{0.314 \text{ mol}}{1.00 \text{ L}} = 0.314 \text{ M}; N_{2}CSH = 0.100 \text{ M}. 3. Reaction with NaOH N_{2}CSH_{2}^+ + OH^- \rightarrow N_{2}CSH + H_2O. Moles of OH⁻ = 0.200 mol. After reaction, N_{2}CSH_{2}^+ = 0.314 - 0.200 = 0.114 \text{ mol}; N_{2}CSH = 0.100 + 0.200 = 0.300 \text{ mol}. 4. Calculate equilibrium concentrations Concentrations: [N_{2}CSH_{2}^+] = \frac{0.114}{1.00} = 0.114 \text{ M}; [N_{2}CSH] = \frac{0.300}{1.00} = 0.300 \text{ M}. 5. Use Henderson-Hasselbalch equation **pH = pK_a + \log\left(\frac A^- HA \right)**. First, find pK_a: pK_a = 14 - pK_b = 14 - (-\log(6.1 \times 10^{-6})) = 8.21. 6. Calculate final pH **pH = 8.21 + \log\left(\frac{0.300}{0.114}\right) = 8.21 + 0.42 = 8.63**.

Explanation

1. Calculate moles of $N_{2}CSH_{2}Cl$<br /> Molar mass of $N_{2}CSH_{2}Cl$ is approximately 95.54 g/mol. Moles = $\frac{30.0 \text{ g}}{95.54 \text{ g/mol}} = 0.314 \text{ mol}$.<br /><br />2. Determine initial concentrations<br /> Initial concentration of $N_{2}CSH_{2}^+$ = $\frac{0.314 \text{ mol}}{1.00 \text{ L}} = 0.314 \text{ M}$; $N_{2}CSH$ = $0.100 \text{ M}$.<br /><br />3. Reaction with NaOH<br /> $N_{2}CSH_{2}^+ + OH^- \rightarrow N_{2}CSH + H_2O$. Moles of OH⁻ = 0.200 mol. After reaction, $N_{2}CSH_{2}^+$ = $0.314 - 0.200 = 0.114 \text{ mol}$; $N_{2}CSH$ = $0.100 + 0.200 = 0.300 \text{ mol}$.<br /><br />4. Calculate equilibrium concentrations<br /> Concentrations: $[N_{2}CSH_{2}^+] = \frac{0.114}{1.00} = 0.114 \text{ M}$; $[N_{2}CSH] = \frac{0.300}{1.00} = 0.300 \text{ M}$.<br /><br />5. Use Henderson-Hasselbalch equation<br /> **pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)**. First, find $pK_a$: $pK_a = 14 - pK_b = 14 - (-\log(6.1 \times 10^{-6})) = 8.21$.<br /><br />6. Calculate final pH<br /> **pH = 8.21 + \log\left(\frac{0.300}{0.114}\right) = 8.21 + 0.42 = 8.63**.
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