Question

Consider a solution formed by mixing 50.0 mL of 0.100 M H2SO4, 30.0 mL of 0.1133 M HOCl, 25.0 mL of 0.200 M NaOH, 25.0 mL of 0.100 M Ba(OH)2, and 10.0 mL of 0.170 M KOH. Calculate the pH of this solution. Ka(HOCl) = 3.5×10-8 pH =

Answer 1

Answer:

The pH of this solution is 12.1.

Explanation:

$H_2SO_4$

$H_2SO_4\rightarrow 2H^++SO_4^{2-}$

1 mole of sulfuric acid gives 2 moles of hydrogen ions, then 0.100 M of sulfuric acid will give :

$[H^+]=2\times [H_2SO_4]=2\times 0.100 M= 0.200 M$

Volume of sulfuric acid solution = 50.0 mL = 0.050 L ( 1 mL = 0.001 L)

Moles of hydrogen ions in sulfuric acid solution :a

$a=0.200 M\times 0.050 L=0.01 mol$

HOCl

The dissociation constant value of HOCl = $K_a=3.5\times 10^{-8}$

$HOCl\rightleftharpoons H^++OCl^-$

initially

0.1133 M         0        0

At equilibrium :

(0.1133-x)         x         x

The expression of dissociation constant is given as:

$K_a=\frac{[H^+][OCl^-]}{[HOCl]}$

$3.5\times 10^{-8}=\frac{x\times x}{(0.1133-x)}$

Solving for x:

x = $6.295\times 10^{-5} M$

$[H^+]=6.295\times 10^{-5} M$

Volume of HOCl solution = 30.0 mL = 0.030 L ( 1 mL = 0.001L)

Moles of hydrogen ions in HOCl solution = b

$b=6.295\times 10^{-5} M\times 0.030 L=1.8885\times 10^{-6} mol$

Total moles of hydrogen ions = a + b

=$0.01 mol +1.8885\times 10^{-6} mol=0.010002 mol$

$NaOH$

$NaOH\rightarrow Na^++OH^{-}$

1 mole of NaOH gives 1 mole of hydroxide ions, then 0.200 M of NaOH acid will give :

$[OH^-]=1\times [NaOH]=1\times 0.200 M= 0.200 M$

Volume of NaOH solution = 25.0 mL = 0.025 L ( 1 mL = 0.001 L)

Moles of hydroxide ions in NaOH solution : c

$c=0.200 M\times 0.025 L=0.005 mol$

$Ba(OH)_2$

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^{-}$

1 mole of $Ba(OH)_2$ gives 2 mole of hydroxide ions, then 0.100 M of

$[OH^-]=2\times [Ba(OH)_2]=1\times 0.200 M= 0.200 M$

Volume of $Ba(OH)_2$solution = 25.0 mL = 0.025 L ( 1 mL = 0.001 L)

Moles of hydroxide ions in $Ba(OH)_2$ solution : d

$d=0.200 M\times 0.025 L=0.005 mol$

$KOH$

$KOH\rightarrow K^++OH^{-}$

1 mole of KOH gives 1 mole of hydroxide ions, then 0.170 M of KOH will give :

$[OH^-]=1\times [KOH]=1\times 0.170 M= 0.170 M$

Volume of KOH solution = 10.0 mL = 0.010 L ( 1 mL = 0.001 L)

Moles of hydroxide ions in KOH solution : e

$e=0.170 M\times 0.010 L=0.0017 mol$

Total moles of hydroxide ions : c + d + e

$0.005 mol + 0.005 mol + 0.0017 mol = 0.0117 mol$

Total moles of hydrogen ions = 0.010002 mol

Total moles of hydroxide ions =  0.0117 mol

1 mole of hydrogen ion reacts with 1 mole of hydroxide ion to from a water with neutral pH.

Hydrogen ions <  hydroxide ion

So, 0.0117 moles of hydroxide ion will neutralize 0.0117 moles of hydrogen ions.

Moles of hydroxide left after neutralization = 0.0117 mol - 0.010002 mol

= 0.001698 moles

Concentration of hydroxide ions left in the solution :

$[OH^-]=\frac{0.001698 mol}{0.050 L+0.030 L+0.025 L+0.025 L+0.010 L}=0.01213 M$

$pOH=-\log[OH^-]=-\log[0.01213 M]=1.9$

pH = 14 - pOH

pH = 14 - 1.9= 12.1

The pH of this solution is 12.1.