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What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?
A
7.0
B
1.04
C
12.65
D
2.0
Detailed Solution
$N_1V_1 - N_2V_2 = NV$: $0.1\times1 - 0.01\times1 = N\times 2$
$[OH^-] = \frac{0.09}{2} = 0.045$ N
pOH = −log(0.045) = 1.35
pH = 14 − 1.35 = 12.65
$[OH^-] = \frac{0.09}{2} = 0.045$ N
pOH = −log(0.045) = 1.35
pH = 14 − 1.35 = 12.65