Question

calculate the pH of a solution formed by mixing 10.0mL of 0.40 M HCl and 20.0 mL of 0.50 M NaOH

Please give as much detail as you can

Answer 1

we have:

M(HCl) = 0.40 M

V(HCl) = 10.0 mL

M(NaOH) = 0.50 M

V(NaOH) = 20.0 mL

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

mol(HCl) = 0.40 M * 10.0 mL = 4.0 mmol

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

mol(NaOH) = 0.50 M * 20.0 mL = 10.0 mmol

We have:

mol(HCl) = 4.0 mmol

mol(NaOH) = 10.0 mmol

4.0 mmol of both will react

remaining mol of NaOH = 6.0 mmol

Total volume = 30.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 6.0 mmol/30.0 mL

= 0.20 M

we have below equation to be used

pOH = -log [OH-]

= -log (0.20)

= 0.699

we have below equation to be used

PH = 14 - pOH

= 14 - 0.699

= 13.301

Answer: 13.3