Respuesta :
Answer:
The pH of this solution is 12.1.
Explanation:
[tex]H_2SO_4[/tex]
[tex]H_2SO_4\rightarrow 2H^++SO_4^{2-}[/tex]
1 mole of sulfuric acid gives 2 moles of hydrogen ions, then 0.100 M of sulfuric acid will give :
[tex][H^+]=2\times [H_2SO_4]=2\times 0.100 M= 0.200 M[/tex]
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
[tex]a=0.200 M\times 0.050 L=0.01 mol[/tex]
HOCl
The dissociation constant value of HOCl = [tex]K_a=3.5\times 10^{-8}[/tex]
[tex]HOCl\rightleftharpoons H^++OCl^-[/tex]
initially
0.1133 M 0 0
At equilibrium :
(0.1133-x) x x
The expression of dissociation constant is given as:
[tex]K_a=\frac{[H^+][OCl^-]}{[HOCl]}[/tex]
[tex]3.5\times 10^{-8}=\frac{x\times x}{(0.1133-x)}[/tex]
Solving for x:
x = [tex]6.295\times 10^{-5} M[/tex]
[tex][H^+]=6.295\times 10^{-5} M[/tex]
Volume of HOCl solution = 30.0 mL = 0.030 L ( 1 mL = 0.001L)
Moles of hydrogen ions in HOCl solution = b
[tex]b=6.295\times 10^{-5} M\times 0.030 L=1.8885\times 10^{-6} mol[/tex]
Total moles of hydrogen ions = a + b
=[tex] 0.01 mol +1.8885\times 10^{-6} mol=0.010002 mol[/tex]
[tex]NaOH[/tex]
[tex]NaOH\rightarrow Na^++OH^{-}[/tex]
1 mole of NaOH gives 1 mole of hydroxide ions, then 0.200 M of NaOH acid will give :
[tex][OH^-]=1\times [NaOH]=1\times 0.200 M= 0.200 M[/tex]
Volume of NaOH solution = 25.0 mL = 0.025 L ( 1 mL = 0.001 L)
Moles of hydroxide ions in NaOH solution : c
[tex]c=0.200 M\times 0.025 L=0.005 mol[/tex]
[tex]Ba(OH)_2[/tex]
[tex]Ba(OH)_2\rightarrow Ba^{2+}+2OH^{-}[/tex]
1 mole of [tex]Ba(OH)_2[/tex] gives 2 mole of hydroxide ions, then 0.100 M of
[tex][OH^-]=2\times [Ba(OH)_2]=1\times 0.200 M= 0.200 M[/tex]
Volume of [tex]Ba(OH)_2[/tex]solution = 25.0 mL = 0.025 L ( 1 mL = 0.001 L)
Moles of hydroxide ions in [tex]Ba(OH)_2[/tex] solution : d
[tex]d=0.200 M\times 0.025 L=0.005 mol[/tex]
[tex]KOH[/tex]
[tex]KOH\rightarrow K^++OH^{-}[/tex]
1 mole of KOH gives 1 mole of hydroxide ions, then 0.170 M of KOH will give :
[tex][OH^-]=1\times [KOH]=1\times 0.170 M= 0.170 M[/tex]
Volume of KOH solution = 10.0 mL = 0.010 L ( 1 mL = 0.001 L)
Moles of hydroxide ions in KOH solution : e
[tex]e=0.170 M\times 0.010 L=0.0017 mol[/tex]
Total moles of hydroxide ions : c + d + e
[tex]0.005 mol + 0.005 mol + 0.0017 mol = 0.0117 mol[/tex]
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 :
[tex][OH^-]=\frac{0.001698 mol}{0.050 L+0.030 L+0.025 L+0.025 L+0.010 L}=0.01213 M[/tex]
[tex]pOH=-\log[OH^-]=-\log[0.01213 M]=1.9[/tex]
pH = 14 - pOH
pH = 14 - 1.9= 12.1
The pH of this solution is 12.1.
