Question

A 25.0 mL sample of 0.150 M potassium hydroxide is titrated with 0.125 M hydrobromic acid solution. Calculate the pH after the following volumes of acid have been added: a) 20.0 mL b) 25.0 mL c) 30.0 mL d) 35.0 mL e) 40.0 mL

Answer 1

Because you have a strong acid and a strong base and the reaction is 1:1 we calculate the moles in each case and the one in excess is the one that is going to give you the pH.

The reaction is NaOH + HCl ----------- NaCl + H2O

So from the NaOH you have moles = Molarity * Volume = 0,150M * 0,025L = 0,00375moles

a) You are adding 20ml of HCl 0,125M

moles = 0,125M* 0,020L= 0,0025moles

Beacuse the reaction is 1:1 0,0025moles of HCl reacts with 0,0025moles of NaOH and the other is the excess of NaOH.

0,00375 - 0,0025 = 0,00125 moles

0,00125 moles/ 0,045L= 0,028M

pOH= -log [0,028] = 5,15

pH = 14-5,15 = 8,85

b) moles= 0,125M*0,025L= 0,0031moles

0,0038moles NaoH - 0,0031moles HCl = 0,0007moles NaOH

0,0007moles/0,050L= 0,014M

pOH= -log [0,014] = 6,15

pH = 14- 6,15 = 7,85

c) moles = 0,125M* 0,030L= 0,00375moles

0,00375moles NaOH - 0,00375 moles HCl =0

[H+] comes from the autoionization of the water where the concentration is 1*10-7M wich is pH =7.

d) moles = 0,125M*0,035L= 0,0044moles

Now the excess has change, the HCl is now the one in excess so is the one that is given the concentration of H+

from the difference

0,0044moles - 0,0038moles= 0,0002moles

0,0002moles/0,060L= 0,0033

pH= -log[H=] = 2,48

e) moles = 0,125M*0,040L =0,005moles

0,005moles- 0,0038 moles= 0,0012moles

0,0012moles/0,065L = 0,018M

pH= -log [0,018]= 1,73