สารละลายกรด HCl pH = 2 ปริมาตร 100 ml เมื่อเติมสารละลายกรด HNO2 pH=1 ปริมาตร 100 ml จะมี pH เท่าใด โดยที่ Ka ของ HNO2 = 4.5x10^-5 (log6.5 = 0.8129) สารละลายกรด HCl pH = 2 ปริมาตร 100 ml เมื่อเติมสารละลายกรด HNO2 pH=1 ปริมาตร 100 ml จะมี pH เท่าใด โดยที่ Ka ของ HNO2 = 4.5x10^-5 (log6.5 = 0.8129) Read more

Steps to Solve

1. Calculate the concentration of $H^+$ from HCl

Given $pH = 2$, we can write the equation as follows:

$[H^+]_{HCl} = 10^{-pH} = 10^{-2} M$

2. Calculate the moles of $H^+$ from HCl

$moles_{HCl} = [H^+]_{HCl} \times Volume = 10^{-2} M \times 100 mL = 10^{-2} M \times 0.1 L = 10^{-3} \text{ moles}$

3. Calculate the concentration of $H^+$ from $HNO_2$

Given $pH = 1$, we can write the equation as follows:

$[H^+]_{HNO_2} = 10^{-pH} = 10^{-1} M$

4. Calculate the moles of $H^+$ from $HNO_2$

$moles_{HNO_2} = [H^+]_{HNO_2} \times Volume = 10^{-1} M \times 100 mL = 10^{-1} M \times 0.1 L = 10^{-2} \text{ moles}$

5. Calculate the total moles of acid ($H^+$) in the mixture

$moles_{total , H^+} = moles_{HCl} + moles_{HNO_2} = 10^{-3} + 10^{-2} = 0.001 + 0.01 = 0.011 \text{ moles}$

6. Calculate the total volume of the solution

$Volume_{total} = 100 , mL + 100 , mL = 200 , mL = 0.2 , L$

7. Calculate the concentration of $H^+$ in the mixture

$[H^+]{mixture} = \frac{moles{total , H^+}}{Volume_{total}} = \frac{0.011}{0.2} = 0.055 , M$

8. Calculate the pH of the mixture

$pH = -\log[H^+]_{mixture} = -\log(0.055)$

$pH = -\log(5.5 \times 10^{-2}) = -(\log 5.5 + \log 10^{-2}) = -(\log 5.5 - 2) = 2 - \log 5.5$

Given $\log 6.5 = 0.8129$, we can estimate $\log 5.5 \approx 0.74$. $pH = 2 - 0.74 = 1.26$ However, the question states that we should take into account the $K_a$ of $HNO_2$. $HNO_2 \rightleftharpoons H^+ + NO_2^-$ Initial: $10^{-1}$ M Change: $-x$ Equilibrium: $10^{-1} - x$ Ignoring the $H^+$ from HCl, the $K_a$ expression is: $K_a = \frac{[H^+][NO_2^-]}{[HNO_2]} \implies 4.5 \times 10^{-5} = \frac{x^2}{0.1 - x} \approx \frac{x^2}{0.1}$ $x^2 = 4.5 \times 10^{-6}$ $x = \sqrt{4.5 \times 10^{-6}} = 2.12 \times 10^{-3}$ Therefore, $[H^+]_{HNO_2} = 2.12 \times 10^{-3}$ upon dissociation. Consider the effect of strong acid added to the weak acid. The equilibrium of the weak acid would be: $HNO_2 \rightleftharpoons H^+ + NO_2^-$ The change in pH would be smaller. Concentration of $HCl = \frac{0.001}{0.2} = 0.005 M$ Concentration of $HNO_2 = \frac{0.01}{0.2} = 0.05 M$ $K_a = \frac{[H^+][NO_2^-]}{[HNO_2]} \implies 4.5 \times 10^{-5} = \frac{(0.005 + x)(x)}{0.05 - x}$ Since $K_a$ is small, approximate $x$ to $0$ $4.5 \times 10^{-5} = \frac{0.005 x}{0.05}$ $x = \frac{4.5 \times 10^{-5} \times 0.05}{0.005} = 4.5 \times 10^{-4}$ Total $H^+ = 0.005 + 4.5 \times 10^{-4} = 0.00545$ $pH = -\log(0.00545) = 2.26$ which is not an option Let's approximate the final pH as determined by HCl only

$moles HCl = 10^{-2}M \times 0.1L = 10^{-3}$ $total volume = 0.2 L$ $[H^+] = \frac{10^{-3}}{0.2} = 5 \times 10^{-3}$ $pH = -\log(5 \times 10^{-3}) = 3 - \log 5 = 3 - 0.6989 = 2.30$ The closest answer to this is 2.19.