Question options:
a) 2.05
b) 0.963
c) 0.955
d) 1.00
Answer:
b) 0.963
Explanation:
H2SO4→ HSO4- + H3O+
HSO4- + H2O ⇌ SO42- + H3O+
Construct ICE table:
HSO4- (aq) + H2O ⇌ SO42- (aq) + H3O+ (aq)
I 0.1 solid & 0 0.1
C -x liquid + x + x
E 0.1 - x are ignored x 0.1 + x
Calculate x
Ka = products/reactants
= [tex]\frac{[SO42-] [H3O+]}{[HSO4-]}[/tex]
0.011 = [tex]\frac{x (0.1 + x)}{0.1 - x}[/tex]
0.011 x (0.1 -x) = o.1x + x^2
0.0011 - 0.011 x - o.1x - x^2 = 0
0.0011 - 0.011 x - x^2 = 0
Use formula to solve for quadratic equation
x = [tex]{ -b +,-\sqrt{b^2 - 4ac[/tex] / 2a
a = -1, b = -0.111, c = 0.001
Solve for x
x = [tex]\sqrt[-(-o.111)]{(-0.111)^2 - 4(-1) (0.0011) }[/tex] / 2(-1)
x = 0.111 +,- [tex]\sqrt{0.012321 + 0.0044}[/tex] / -2
x = 0.111 +,- [tex]\sqrt{0.016721}[/tex] / -2
x = [tex]\frac{0.111 +, - 0.1293}{-2}[/tex]
x = [tex]\frac{0.111 + 0.1293}{-2}[/tex] , x = [tex]\frac{0.111 - 0.1293}{-2}[/tex]
x = [tex]\frac{0.2403}{-2}[/tex] , x = [tex]\frac{0.0183}{-2}[/tex]
x = - 0.12015 , x = 0.00915
x cannot be negative, so
x = 0.00915 M
Calculate [H3O+]
[H3O+] = 0.1 M + x
[H3O+] = 0.1 M + 0.00915 M
[H3O+] = 0.10915 M
Clculate pH
pH = - log [ H3O+]
pH = - log [ 0.10915]
pH = 0.963
