Question

On adding 0.04 g solid NaOH to a 100 mL, M/200. Ba(OH)2 solution, determine change in pH :
(A) O
(B) +0.3
(C) -0.3
(D) +0.7​

Answer 1

${\huge\bf{\mid{\overline{\underline{+0.3}}}\mid}}$

Answer 2

The correct answer is option B.

Explanation:

$Molarity=\frac{Moles}{Volume(L)}$

Moles of barium hydroxide solution =  n

Molarity of the solution = M/200 = 0.005 M

Volume of the solution = V = 100 ml - 0.1 L( 1mL = 0.001 L)

$n=0.005 M\times 0.1 M = 0.0005 mol$

1 mole of barium hydroxide gives 2 moles of moles of hydroxide ions.Then 0.0005 mole of barium hydroxide will give:

$0.0005 mol\times 2=0.001 mol$ hydroxide ions

Concentration of hydroxide ions ;

$=\frac{0.001 mol}{0.1 L}=0.01 M$

$pOH=\log[OH^-]=-\log [0.01]=2$

pH = 14 - pOH = 14 - 2 = 12

Initial pH of the solution = 12

Th pH of the after adding 0.04 grams of sodium hydroxide:

Moles of NaOH = $\frac{0.04 g}{40 g/mol}=0.001 mol$

1 mole of NaOH has 1 mole of hydroxide ions. Then 0.001 mole of NaOH will have :

$1\times 0.001 mol=0.001 mol$ of hydroxide ions

Total moles of hydroxide ions in the solution = 0.001 mol + 0.001 mol= 0.002 mol

Concentration of hydroxide ions ;

$=\frac{0.002 mol}{0.1 L}=0.02 M$

$pOH'=-\log[0.02]=1.7$

$pH'=14-1.7=12.3$

final pH of the solution = 12.3

Change in pH = 12.3 - 12 = 0.3

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