Question

Consider the titration of a 38.0mL sample of 0.170M  HBr with 0.205M KOH. Determine each of the following:

Part A: the initial pH

Part B: the volume of added base required to reach the equivalence point

Part C: the pH at 11.0mL of added base

Part D: the pH at the equivalence point

Part E: the pH after adding 5.0 mL of base beyond the equivalence point

Answer 1

Both HBr and KOH are strong acid and strong base hence they dissocite completely to give H+ and OH- ions respectively.

[HBr] = 0.170 M

[KOH] = 0.205 M

Part A:

initial [H+] = 0.17 M

pH = - log [H+]

= - log ( 0.17)

= 0.77

Part B:

at equivalence point moles of HBr = moles of KOH

38 * 0.17 = V * 0.205

Vol of KOH in ml = 31.51 ml

Part C:

at 11.0mL of added base

excess H+ moles = ( 0.17 * 38/1000) - ( 0.205 * 11.0/1000)

= 0.004205

total volume = 11 + 38 = 49 ml

= 0.049 L ,

Total [H+] = ( 0.004205 / 0.049)

= 0.0858 M

pH = - log ( 0.0858)

= 1.06

Part D:

at equivalence point

H+ moles = OH- moles

hence solution is neutral and hence

pH = 7

Part E:

the pH after adding 5.0 mL of base beyond the equivalence point

excess moles of OH- = 0.005*0.205

= 0.001025

total volume = 38.0 + 31.51 + 5.0 = 74.51 mL

= 0.07451 L

[OH-] = 0.001025 / 0.07451

= 0.01375 M

pOH = - log[OH-]

= - log(0.01375)

= 1.861

pH = 14 - 1.861

= 12.139