Question

How many units of pH increase will occur if 10.0 mL of 0.010 M NaOH is added to 90.0 mL of pure water?

when some solid potassium fluoride is added to a hydrofluoric acid solution, will the pH of the solution be higher, lower, or the same?

Answer 1

pH of pure water is 7

now as NaOH is a base, when added to water increases its pH

we know that pH = -log[H⁺]

pOH = -log[OH^-]

and 14 = pH + pOH

lets calculate pOH first,

volume = 10 mL = 0.01L

Moles of NaOH = molarity * volume in litres = 0.010 M * 0.01 L = 10^-4 moles

total volume = 10 mL(NaOH) + 90 mL(water) = 100 mL

= 0.1 L

now molar concentration of [OH^-] = [NaOH] as it will completly dissociate in water as

NaOH(aq) = Na⁺ (aq)+ OH^-(aq)

[OH^-] = 10^-4 moles / 0.1 L = 10^-3 M

pOH = -log[10^-3] = 3

now pH = 14-pOH = 14 - 3 = 11

no of units of pH increased = 11-7 =4 units

when KF is added to HF its acidity will decrease due to common ion effect

HF(aq) = H⁺(aq) + F^-(aq)

now KF(aq) = K⁺(aq) + F^-(aq) here F^- dissociated from this will increase the overall conc. of F^- in the solution which will suppress the dissociation of HF to H⁺ (which is reason for acidity here) can also be understood by using le chateliers principle. pH of solution will be higher as more is the concentraion of H⁺ more is the acidity and less is pH