Question

25 ml of 0.105 M HCl is titrated with 0.210 M NaOH

a. What is the ph after 5 ml of the base is added?

b. what is the ph at the equivalence point?

c. What is the ph after 15 ml of the base added?

d. how many ml of the base will be required to reach the end point?

Answer 1

25ml of 0.105 M HCl is titrated with 0.210 M NaOH

a). What is the ph after 5 ml of the base is added?

Solution :-

moles of HCl = mol per L * 0.025 L = 0.002625 mol

Moles of NaOH = 0.210 mol per L * 0.005 L = 0.00105 mol

Moles of NaOH are less than moles of HCl therefore

Moles of HCl remain after reaction = 0.002625 mol -0.00105 mol = 0.001575 mol HCl

Total volume = 25 ml + 5 ml = 30 ml = 0.030 L

New molarity of the HCl = 0.001575 mol / 0.030 L = 0.0525 M

pH= -log [H+]

pH= -log [0.0525]

pH= 1.28

b. what is the ph at the equivalence point?

Solution :-

At the equivalence point moles of the acid and base that is moles of HCl and moles of NaOH are same therefore they completely neutralize each other

So the solution becomes neutral that means the pH is 7.00

c. What is the ph after 15 ml of the base added?

Solution :-

Moles of NaOH = 0.210 mol per L * 0.015 L = 0.00315 mol

moles of HCl = mol per L * 0.025 L = 0.002625 mol

here moles of NaOH are more than moles of HCl

therefore moles of NaOH remain after reaction = 0.00315 mol – 0.002625 mol = 0.000525 mol NaOH

total volume = 25 ml + 15 ml = 40 ml = 0.040 L

new molarity of the NaOH = 0.000525 mol / 0.040 L = 0.013125 M

pOH = -log [OH-]

pOH= - log [ 0.013125]

pOH = 1.88

pH + pOH = 14

pH = 14 – pOH

pH= 14 – 1.88

pH= 12.12

d. how many ml of the base will be required to reach the end point?

Solution :-

Mole ratio of the NaOH and HCl is 1 : 1

So lets calculate the volume of NaOH needed to reach the equivalence point

Volume of NaOH = molarity of HCl * volume of HCl / molarity of NaOH

                              = 0.105 M * 25 ml / 0.210 M

                              = 12.5 ml

So to reach the end point means we need little excess NaOH means it can be 12.6 ml.