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What will be the change in the pH of wat...

What will be the change in the pH of water if `10^(-3)` mol of NaOH is added to `1.0` L of water ?

A

Decreased by 4

B

Increased by 4

C

Increased by 3

D

Decreased by 4

Text Solution

Verified by Experts

The correct Answer is:
B
Before adding NaOH, water is nautral and has pH `= 7` at `25^(@)C`.
`C_(NaOH)=(n_(NaOH))/(V_(L)) = (10^(-3)mol)/(1.0L)= 10^(-3)mol L^(-1)`
Since NaOH is a strong base and 1 mol NaOH gives 1 mol `OH^(-)`,
`C_(OH^(-))=C_(NaOH)=10^(-3)M`
Comparing it with the expression
`C_(OH^(-))=10^(-pOH)M`
we get pOH = 3
At `25^(@)C`,
`pK_(w)=pH+pOH=14`
`:. pH = 14 - pOH = 14-3 = 11`
Change in pH = (Final pH) - (Initial pH)
`= 11-7`
`= 4`
i.e., pH increases by 4.
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