Consider the titration of a 25.1 −mL sample of 0.125 M RbOH with 0.100 M HCl. Determine each of the following.the initial pH, the volume of added acid required to reach the equivalence point,he pH at 4.9 mL of added acid,the pH at the equivalence pointthe pH after adding 4.2 mL of acid beyond the equivalence point
RbOH + HCl ===⇒ RbCl + H2O
A.
[H+][OH-] = 1.0 x 10^-14
[H+] = 1.0 x 10^-14 / 0.125 = 8 x 10^-14
pH = -log(8 x 10^-14 ) = 13.1
B.
25.1 mL x 0.125 moles RbOH/1liter x 1liter/1000 mL = 0.003137 moles
RbOH
0.003137 moles RbOH x 1 mole HCl/1mole RbOH x 1 liter HCl/0.10
moles HCl = 0.03137 liters or 31.37
mL
C.
4.9 mL of HCl x 0.10 moles HCl/ 1 liter x 1 liter/1000ml x 1 mole
RbOH/1mole HCl = 0.00049 moles RbOH consumed.
Mole of RbOH remaining = 0.003137 – 0.00049 = 0.002647
Molarity OH- = 0.002647 moles / 25.1 mL + 4.9 mL x 1000
mL /liter = 0.0882 M
[H+]OH-] = 1.0 x 10^-14
[H+] = 1.0 x 10^-14 /0.0882 = 1.13 x 10^-13
pH = - log(1.13 x 10^-13) = 12.94
D. neutral solution: pH =
7.0
E. Final volume of solution = 4.2 mL +
31.37 mL(from part B). = 35.57 mL
Moles of HCl added after equivalence point.
4.2 mL x 0.10 moles HCl / 1liter x 1 liter/1000 mL = 0.00042
moles
Molarity of [H+] = 0.00042 moles/ 35.57 mL x 1000 mL/1 liter =
0.0118
pH = -log( 0.0118) = 1.93
Consider the titration of a 25.1 −mL sample of 0.125 M RbOH with 0.100 M HCl....
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