Question

Calculate the pH of the resulting solution if 25.0 mL of 0.250 M HCl(aq) is added to 35.0 mL of 0.250 M NaOH(aq).

pH=

Calculate the pH of the resulting solution if 25.0 mL

of 0.250 M HCl(aq) is added to 15.0 mL of 0.350 M NaOH(aq).

pH=

No referals as the only answer unless you're the first one.

Answer 1

1)

Given:

M(HCl) = 0.25 M

V(HCl) = 25 mL

M(NaOH) = 0.25 M

V(NaOH) = 35 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.25 M * 25 mL = 6.25 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.25 M * 35 mL = 8.75 mmol

We have:

mol(HCl) = 6.25 mmol

mol(NaOH) = 8.75 mmol

6.25 mmol of both will react

remaining mol of NaOH = 2.5 mmol

Total volume = 60.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 2.5 mmol/60.0 mL

= 4.167*10^-2 M

use:

pOH = -log [OH-]

= -log (4.167*10^-2)

= 1.3802

use:

PH = 14 - pOH

= 14 - 1.3802

= 12.6198

Answer: 12.62

2)

Given:

M(HCl) = 0.25 M

V(HCl) = 25 mL

M(NaOH) = 0.35 M

V(NaOH) = 15 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.25 M * 25 mL = 6.25 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.35 M * 15 mL = 5.25 mmol

We have:

mol(HCl) = 6.25 mmol

mol(NaOH) = 5.25 mmol

5.25 mmol of both will react

remaining mol of HCl = 1 mmol

Total volume = 40.0 mL

[H+]= mol of acid remaining / volume

[H+] = 1 mmol/40.0 mL

= 2.5*10^-2 M

use:

pH = -log [H+]

= -log (2.5*10^-2)

= 1.6021

Answer: 1.60