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# Calculate the pH of the resultant mixtures: a) 10 mL of 0.2M Ca(OH)_2 + 25 mL of 0.1M HCl

7.66     Calculate the pH of the resultant mixtures:

a)       10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl

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Given that,
Vol. of 0.2 M $Ca(OH)_2$ = 10 mL
Vol. of 0.1 M HCl = 25 mL

therefore, by using the formula,

$M(OH^-)= \frac{M_1V_1(base)-M_2V_2(acid)}{V_1+V_2}$
By substituting the value in these equations, we get;

$\\\Rightarrow \frac{(0.2 \times 2)- (0.1\times 2)}{10+25}\\ =\frac{1.5}{25}=0.06$

Now, $p^{OH} = -\log[OH^-]$
$\\= -\log(0.06)\\ =1.221$

since $p^H+p^{OH} = 14$
$p^H=14-p^{OH}$
= 14-1.221
=   12.78

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