Search any question & find its solution
Question:
Answered & Verified by Expert
The pH of a $10^{-10} \mathrm{M}-\mathrm{NaOH}$ solution is
Options:
Solution:
2105 Upvotes
Verified Answer
The correct answer is:
7.01
When the solution is very dilute, the concentration of $\mathrm{OH}^{-}$produced from water cannot be neglected. Hence,
$\begin{aligned} & {\left[\mathrm{OH}^{-}\right]=10^{-10}+10^{-7}} \\ & \text { (obtained from water) } \\ & =10^{-7}(0.001+1)=1.001 \times 10^{-7} \\ & \therefore \quad \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right] \\ & =-\log \left(1.001 \times 10^{-7}\right) \\ & =7-0.01=6.99 \\ & \therefore \quad \mathrm{pH}=14-\mathrm{pOH} \\ & =14-6.99=7.01 \\ & \end{aligned}$
$\begin{aligned} & {\left[\mathrm{OH}^{-}\right]=10^{-10}+10^{-7}} \\ & \text { (obtained from water) } \\ & =10^{-7}(0.001+1)=1.001 \times 10^{-7} \\ & \therefore \quad \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right] \\ & =-\log \left(1.001 \times 10^{-7}\right) \\ & =7-0.01=6.99 \\ & \therefore \quad \mathrm{pH}=14-\mathrm{pOH} \\ & =14-6.99=7.01 \\ & \end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.