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Chemistry Question on Equilibrium

Assuming that the degree of hydrolysis is small, the pH of 0.1 M solution of sodium acetate (Ka=1.0×10−5)(K_a = 1.0 \times 10^{-5}) will be :

A

5

B

6

C

8

D

9

Answer

9

Explanation

Solution


⇒Kh=[CH3COOH][H?][CH3COO]=KwKa=ch2(1−h)\Rightarrow K _{ h }=\frac{\left[ CH _{3} COOH \right][ \overset{?}{ H} ]}{\left[ CH _{3} COO \right]}=\frac{ K _{ w }}{ K _{ a }}=\frac{ ch ^{2}}{(1- h )}
⇒10−1410−5=ch2\Rightarrow \frac{10^{-14}}{10^{-5}}= ch ^{2}
\\{\because h is very small \therefore 1- h \approx 1\\}
⇒h=10−90.1=10−4\Rightarrow h =\sqrt{\frac{10^{-9}}{0.1}}=10^{-4}
∴[O?H]=ch=0.1×10−4=10−5\therefore[ \overset{?}{O}H ]= ch =0.1 \times 10^{-4}=10^{-5}
⇒[H⊕]=10−9\Rightarrow[ \overset{\oplus}{H }]=10^{-9}
∴pH=−log⁡[H⊕]=9\therefore p ^{ H }=-\log [ \overset{\oplus}{H} ]=9