Question

Calculate the pH of the resulting solution if 35.0 mL of 0.350 M HCl(aq) is added to

(a) 45.0 mL of 0.350 M NaOH(aq).

pH=

(b) 25.0 mL of 0.450 M NaOH(aq).

pH=

Answer 1

1) The first step you do is convert everything to moles. You are given concentration and volume, so you can find moles by the equation:

concentration * volume = moles

35.0 mL * 0.35M HCl = 12.25mmoles HCl

45.0 mL * 0.350 M NaOH = 15.75 mmoles NaOH

You know that HCl and NaOH neutralize each other by the reaction:

HCl + NaOH ---> NaCl + H2O

So subtract the two mmole amount from each other. The larger value will be the excess of that amount.

15.75 mmoles NaOH- 12.25 mmoles HCl = 3.5 mmoles NaOH leftover

Now you have to find concentration again, which you can find by:

moles / total volume = concentration

3.5 mmoles / (35mL + 45mL) = 0.04375 M NaOH

Now you can find the pOH, which is -log[OH-].

pOH = -log(0.04375) = 1.359

so pH = 14- pOH = 14-1.359 = 12.641 .............answer

2)

same as above

concentration * volume = moles

35.0 mL * 0.35M HCl = 12.25mmoles HCl

25.0 mL * 0.45M NaOH = 11.25 mmoles NaOH

You know that HCl and NaOH neutralize each other by the reaction:

HCl + NaOH ---> NaCl + H2O

So subtract the two mmole amount from each other. The larger value will be the excess of that amount.

12.5 mmoles HCl- 11.25 mmoles NaOH = 1.25 mmoles HCl leftover

Now you have to find concentration again, which you can find by:

moles / total volume = concentration

1.25 mmoles / (35mL + 45mL) = 0.015625 M HCl

Now you can find the pH, which is -log[H+].

pH = -log(0.015625) = 1.806 ............answer