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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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M manish

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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