Q. A solution of $\ce{HCl}$ has a $\mathrm{pH}$ of $5$. If $1\ \mathrm{ml}$ is diluted to one litre of water at $\mathrm{pH}=7$. What is the $\mathrm{pH}$ of the resulting solution.


$$\begin{align} \mathrm{pH}&=-\log[\ce{H+}]=5\\ [\ce{H+}]&=10^{-5}\\ M_1V_1&=M_2V_2\\ 10^{-5}\times10^{-3}&=M_2\times1\\ M_2 &= 10^{-8}\end{align}$$

Since $[\ce{H+}]<10^{-7}$ Thus, by common ion effect, $$\begin{align}[\ce{H+}]&=10^{-7}[1+0.1]\\ [\ce{H+}]&=1.26 \times 10^{-6}\\ \mathrm{pH}&=-[\log(1.26)+(-6)]=5.9 \end{align}$$

But, the answer sheet mentions it to be 6.97.


Everything you did is correct except a little math error. When you wrote:

$$[\ce{H+}] = 10^{-7}[1+0.1]\\ [\ce{H^{+}}] = 1.26 \times 10^{-6}$$

it should instead be

$$[\ce{H+}] = 10^{-7}[1+0.1]\\ [\ce{H^{+}}] = 1.1 \times 10^{-7}$$

and thus

$$\mathrm{pH} =-\log(1.1\times 10^{-7}) = 6.96$$

  • 1
    $\begingroup$ For some reason I decided to take antilog of 1.01. Thanks for pointing it out. $\endgroup$ – Akshat Batra Nov 14 '16 at 6:19

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