A What is the pH of 0040 M HF B What is the pH of 0035 M eth

A) What is the pH of 0.040 M HF? B) What is the pH of 0.035 M ethylammonium chloride, C2HsNHsCI? C) What is the pH of0.540 M rubidium perchlorate, RbCIO? D) What is the pH of 0.13 M sulfurous acid, H2SO3? E) What is the pH of 0.060 M sodium oxalate, Na2C204? 3.

Solution

pH = -log[H+] for strong acids

pOH = -log[OH-] for strong bases

pH = 1/2(pKa - logC) for weak acid

pOH = 1/2(pKb - logC) for weak base

a) HF is weak acid so Ka = 6.6 * 10^-4

pH = 1/2(-log(6.6*10^-4) - log(0.04))

pH = 2.2892

b) C2H5NH3Cl is salt of weak base C2H5NH2 and strong acid HCl

pH = 7 - 1/2(pKb + logC) ( C = concentration of the salt)

pH = 7 - 1/2(-log(5.6*10^-4) + log(0.035))

pH = 6.102

c) RbClO4 is a salt of strong base RbOH and strong acid HClO4 hence the salit will be neutral.

so pH = 7 for that solution.

d)

H2SO3 is a moderate acid acid so,

pH = 1/2(pKa - logC)

pH = 1/2(-log(1.5*10^-2) - log(0.13))

pH = 1.355