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7.49     Calculate the pH of the following solutions:

 (d)      1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.

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We know that,
M1V1(before dilution) = M2V2(after dilution)

initially V1 = 1mL and M= 13.6 M
and V2 = 1L and M= ?

By putting all these values we get,

M_2 = \frac{13.6\times 10^{-3}}{1}=1.36\times 10^{-2}
\therefore [H^+]=1.36\times 10^{-2}

Thus P^H = -\log [H^+] = -\log(1.36\times 10^{-2})
                                               = 1.86 (approx)

 

Posted by

manish

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