In 20 , mL of 0.4 , M HA solution,80 , mL of | vedclass

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(B) Initial moles of $HA = 20 \, mL 	imes 0.4 \, M = 8 \, mmol$.Total volume of the final solution = $20 \, mL + 80 \, mL = 100 \, mL$.Final concentr

Vedclass · Accepted Answer

$3.75$

Vedclass · Answer

(B) Initial moles of $HA = 20 \, mL 	imes 0.4 \, M = 8 \, mmol$.Total volume of the final solution = $20 \, mL + 80 \, mL = 100 \, mL$.Final concentration of $HA$ $(C)$ = $\frac{8 \, mmol}{100 \, mL} = 0.08 \, M$.For a weak acid,$[H^+] = \sqrt{K_a \cdot C} = \sqrt{4 	imes 10^{-7} 	imes 0.08} = \sqrt{32 	imes 10^{-9}} = \sqrt{3.2 	imes 10^{-8}} \approx 1.788 	imes 10^{-4} \, M$.Alternatively,$[H^+] = \sqrt{32 	imes 10^{-9}} = \sqrt{16 	imes 2 	imes 10^{-10} 	imes 10} = 4 	imes 10^{-5} 	imes \sqrt{2} \approx 4 	imes 1.414 	imes 10^{-5} = 5.656 	imes 10^{-5} \, M$.Wait,let's re-calculate: $[H^+] = \sqrt{32 	imes 10^{-9}} = \sqrt{320 	imes 10^{-10}} = 17.88 	imes 10^{-5} = 1.788 	imes 10^{-4} \, M$.$pH = -\log(1.788 	imes 10^{-4}) = 4 - \log(1.788) \approx 4 - 0.25 = 3.75$.

In $20 \, mL$ of $0.4 \, M$ $HA$ solution,$80 \, mL$ of water is added. Assuming volume to be additive,the $pH$ of the final solution is ($K_a$ of $HA = 4 \times 10^{-7}, \log 2 = 0.3$).

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