Question

Calculate the pH N/1000 NaOH solution assuming complete ionisation???

Answer 1

[OH-] = Normality = 10^-3

pOH = -log[OH-]
= -log(10^-3)
= 3

pH = 14 - pOH
= 14 - 3
= 11

pH of N/10 NaOH solution is 11

Answer 2

Given:

$\frac{N}{1000}$ $NaOH$

To find

pH = ?

Formula to be used:

$pOH=-log_{10}[OH^-]$

$pH=14-pOH$

Calculation:

As NaOH is monoacidic base, for it

Molarity = Normality

$[OH^-]$$= [NaOH]=\frac{N}{1000} = 0.001N = 0.001M$

$pOH=-log_{10}[OH^-]$

$pOH=-log_{10}[0.001]$

$pOH= 3$

Substituting pOH value in the following formula we get,

$pH=14-pOH$

$pH=14-3$

$pH=11$

Conclusion:

The pH of N/1000 NaOH solution upon complete ionization is 11.

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