Question

0.023 g of sodium metal is reacted with 100 cm3 of water. The pH of the resulting solution is ______.

(A) 11
(B) 10
(C) 12
(D) 9

Answer 1

Na + H2O => NaOH 

[NaOH] = (0.023/23)/0.1 = 0.01 M 
so 
pH = 12

(C)

Answer 2

Given:

Weight of sodium metal = 0.023g

Volume of water = 100$cm^3$ = $10^{-1}$ liters

To find:

pH of the solution =?

Formula to be used:

pH = $-log[H^+]$

Calculation:

No. of moles = $\frac{Given mass}{Molar mass}$ = $\frac{0.023}{23}= 10^{-3}$

$[OH^-] = \frac{No. of moles }{Volume}$

$[OH^-]=\frac{10^{-3}}{10^{-1}}$

$OH^- = 10^{-2}$

WKT

$[H^+][OH^-] = 10^{-14}$

∴$H^+$ = $10^{-12}$

pH = $-log[H^+]$

pH = -log ($10^{-12}$)

pH = 12

Conclusion:

The pH of the resulting solution is calculated as 12.

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