Question

1. calculate the pH of a solution with [H3O+]= 2.4x10^-5M
2. the pOH for a KOH solution is 4.5, determine [H+] for the solution.
3. A 15.00 mL sample of NaOH solution of unknown concentration required 17.88 mL of 0.1053 M H2SO4 solution to reach the equivalent point in a titratiom. what is the concentration of the NaPH Solution?

Answer 1

1) pH of strong acid or stong base = -log[H+] or -log{OH-]

[H3O+] is [H+] { H+ is concentration }

pH = -log[H3O+]

pH = -log [2.4 x 10^-5] = 4.62

2) KOH ----> K⁺ + OH^-

molar ration of K+ and OH- is 1:1 so

[OH-] = [K+]

pOH = -log[OH-]

4.5 = -log[OH-]

[OH-] = 10^-4.5 = 3.16 x10^-5 M

[K+] = [H+] = 3.16 x10^-5 M

3) Write the chemical equation for the reaction:

2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O

The key molar ratio is that of H₂SO₄ to NaOH, is 1 : 2

Determine moles of H₂SO₄ :

moles = MV = (0.1053 mol/L) (0.01788 L) = 0.00188 mol

Determine moles of NaOH:

ratio is 1 : 2

0.00188 x 2 mol = X

X = 0.00376 mol of NaOH consumed

Determine molarity of NaOH solution:

0.00376 mol / 0.015 L = 0.25 M