Question

Consider the titration of 35.0 ml sample of 0.175 M Her with 0.200 M KOH. Determine the quantities below:
Equation of reaction:HBr(aq)+KOH(aq)<>KBr(aq)+H2O(l)
a) initial oh, before the addition of base
b) ph at 10.00ml added base
c) pH at equivalent point
d) ph after adding 35.00 ml of base

Answer 1

a) initial oh, before the addition of base

pH of the solution = -log 0.175 = 0.756

b) ph at 10.00ml added base

Moles of HBr present before addition = 35 x 0.175 / 1000 = 0.006125 Moles

Moles of base added = 10 x 0.2 /1000 = 0.002 Moles

Moles of HBr present after addition = 0.006125 - 0.002 = 0.004125 Moles

Volume of the solution = 35 + 10 = 45 ml

concentration of the solution = 0.004125 x1000 /45 = 0.09166

pH of the solution = - log 0.09166 = 1.03

c) pH at equivalent point

Moles of base added = 10 x 0.2 /1000 = 0.004125 Moles

Moles of HBr present after addition = 0.004125 - 0.004125 =  0 Moles

pH of the solution = 7

d) ph after adding 35.00 ml of base

Moles of HBr present before addition = 35 x 0.175 / 1000 = 0.006125 Moles

Moles of base added = 35 x 0.2 /1000 = 0.007 Moles

Moles of KOH present after addition = 0.007 - 0.006125 =  0.000875 Moles

Volume of the solution = 35 + 35 = 70 ml

concentration of the solution = 0.000875 x1000 /70 = 0.0125

pOH of the solution = - log 0.0125= 1.903

pH of the soluiotn = 40-1.903 = 12.096