First calculate volume of HCl required to reach the equivalence point
Consider reaction, HCl + KOH KCl + H2O
From reaction, 1 mol HCl 1 mol KOH
We have relation, M acid V acid = M base V base
V acid = M base V base / M acid
V acid = 0.156 M 50.0 ml / 0.312 M = 25.0 ml
Volume of acid required to reach equivalence point = 25.0 ml
pH after addition of 20.0 ml HCl
mmol of HCl = concentration volume = 0.312 20.0 = 6.24 mmol
mmol of KOH = 0.156 50.0 = 7.8 mmol
mmol of excess KOH = 7.8 - 6.24 = 1.56 mmol
Volume of mixture at this stage = volume of HCl + volume of KOH = 20.0 + 50.0 = 70.0 ml
[KOH] = No. of moles of KOH / volume of solution in L
[KOH] = 1.56 /1000] / [70.0 /1000]
[KOH] = 0.02228 M
We have , pOH = -log [OH- ] = - log 0.02228 = 1.67
We have relation, pH + pOH = 14
pH = 14 - pOH = 14 -1.67 = 12.33
pH after addition of 24.0 ml HCl
mmol of HCl = concentration volume = 0.312 24.0 = 7.488 mmol
mmol of KOH = 0.156 50.0 = 7.8 mmol
mmol of excess KOH = 7.8 -7.488 = 0312 mmol
Volume of mixture at this stage = volume of HCl + volume of KOH = 24.0 + 50.0 = 74.0 ml
[KOH] = No. of moles of KOH / volume of solution in L
[KOH] =[ 0.312/ 1000] / [74.0 /1000]
[KOH] = 0.00422 M
We have , pOH = -log [OH- ] = - log 0.00422 =2.37
We have relation, pH + pOH = 14
pH = 14 - pOH = 14 -2.37 = 11.63
pH after addition of 25.0 ml HCl (At equivalence point)
At equivalence point, all HCl is consumed by added KOH, solution contain salt KCl. Salt does not undergo hydrolysis, hence pH of solution will be due to dissociation of water. Therefore, pH will be 7.0
pH after addition of 26.0 ml HCl
After equivalence point, there is excess HCl and pH of solution depends on concentration of HCl.
mmol of HCl = 0.312 26.0 = 8.112 mmol
mmol of KOH = 0.156 50.0 = 7.8 mmol
mmol of excess HCl = 8.112 - 7.8 = 0.312
Volume of solution at this stage = volume of HCl + volume of KOH = 26.0 + 50.0 = 76.0 ml
[HCl] = [ 0.312 /1000] /[ 76.0 /1000] = 0.00411 M
We have, pH = -log [H +] = - log 0.00411 = 2.39
pH after addition of 30.0 ml HCl
mmol of HCl = 0.312 30.0 =9.36 mmol
mmol of KOH = 0.156 50.0 = 7.8 mmol
mmol of excess HCl = 9.36 - 7.8 =1.56
Volume of solution at this stage = volume of HCl + volume of KOH = 30.0 + 50.0 =80.0 ml
[HCl] = [ 1.56 /1000] /[ 80.0/1000] = 0.0195 M
We have, pH = -log [H +] = - log 0.0195 =1.73
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