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Chemistry Question on Equilibrium

The pH{pH} of a 10−10 M NaOH{10^{-10}\, M\, NaOH} solution is

A

10

B

7.1

C

6.99

D

4

Answer

7.1

Explanation

Solution

When the solution is very dilute, the concentration of OH−{OH^-} produced from water cannot be neglected. Hence,
[OH−]=10−10+10−7{ [OH^-] = 10^{-10} + 10^{-7} } (obtained from water)
=10−7(0.001+1)= 10^{-7} (0.001 + 1)
=1.001×10−7= 1.001 \times 10^{-7}
∴    pOH=−log[OH−]\therefore \:\:\:\: {pOH = - log[OH^-]}
=−log(1.001×10−7)=6.99{= - log(1.001 \times 10^{-7}) = 6.99}
∴   pH=14−pOH\therefore \:\:\: {pH = 14 - p OH}
14−6.99=7.0114 - 6.99 = 7.01