Math  /  Algebra

Question1. What is the pH of a buffer solution prepared by adding 2.12×103 mL2.12 \times 10^{3} \mathrm{~mL} of 0.028MCa(OH)20.028 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2} to 242 mL of 0.819 M HF ?
Note: the pKap K_{a} of HF is 3.20 , you may assume the 5%5 \% approximation holds, and you may assume that the volumes are additive. INSTRUCTIONS: Input your answer to 2 decimal places in standard notation (example: 1.23 ) and DO NOT include the units. pH=\mathrm{pH}= \square Submit Answer Tries 0/5

Studdy Solution
Use the Henderson-Hasselbalch equation to find the pH.
The Henderson-Hasselbalch equation is:
pH=pKa+log([A][HA]) \mathrm{pH} = pK_a + \log \left( \frac{[\mathrm{A}^-]}{[\mathrm{HA}]} \right)
Where [A] [\mathrm{A}^-] is the concentration of F \mathrm{F}^- and [HA] [\mathrm{HA}] is the concentration of HF \mathrm{HF} .
Total volume = 2.12L+0.242L=2.362L 2.12 \, \mathrm{L} + 0.242 \, \mathrm{L} = 2.362 \, \mathrm{L} .
[F]=0.11872mol2.362L=0.05026M [\mathrm{F}^-] = \frac{0.11872 \, \mathrm{mol}}{2.362 \, \mathrm{L}} = 0.05026 \, \mathrm{M}
[HF]=0.079478mol2.362L=0.03365M [\mathrm{HF}] = \frac{0.079478 \, \mathrm{mol}}{2.362 \, \mathrm{L}} = 0.03365 \, \mathrm{M}
pH=3.20+log(0.050260.03365) \mathrm{pH} = 3.20 + \log \left( \frac{0.05026}{0.03365} \right)
pH=3.20+log(1.494) \mathrm{pH} = 3.20 + \log (1.494)
pH=3.20+0.174 \mathrm{pH} = 3.20 + 0.174
pH=3.37 \mathrm{pH} = 3.37
The pH of the buffer solution is:
3.37 \boxed{3.37}

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