Question

What is the pH of a solution in which 15 mL of 0.59 M NaOH is added to 25 mL of 0.59 M HCl?

Answer 1

Since NaOH is a strong base and HCl is a strong acid, they react completely

Thus, HCl(aq) + NaOH(aq) ----> NaCl(aq) + H₂O(l)

moles of NaOH = molarity*volume of solution in litres = 0.59*0.015 = 0.00885

moles of HCl = molarity*volume of solution in litres = 0.59*0.025 = 0.01475

Clearly, NaOH is limiting

Thus, HCl unreacted = 0.0059

[HCl] unreacted = moles/volume of solution in litres = 0.0059/0.04 = 0.1475 M

Now, HCl(aq) ----> H⁺(aq) + Cl^-(aq)

Thus, [H⁺] = 0.1475 M

pH = -log[H⁺] = 0.831