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# Calculate the pH of the resultant mixtures: c) 10 mL of 0.1M H_2SO_4 + 10 mL of 0.1M KOH

7.66     Calculate the pH of the resultant mixtures:

c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH

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Given that,

Volume of 0.1 M $KOH$ = 10 mL, and

Volume of 0.1 M $H_2SO_{4}$ = 10 mL

So, by using the formula of,

$M(H^+)=\frac{M_1V_1(acid)-M_2V_2(base)}{V_1+V_2}$
By putting the values we get,

$\\\Rightarrow \frac{2(0.1\times 10)-0.1\times 10}{10+10}\\ =\frac{1}{20}\\ =5\times 10^{-2}$

Hence, $pH = -\log[H^+] = -\log(5\times 10^{-2})= 1.30$

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