Question

What is the pH of 5.0756mg barium hydroxide octahydrate in4.6 liters of water, and the pH of 0.19789g dihydrogen sulfate in 374mL of water both are separate and at 25 degrees C with nothing else present, assume no change of volume and or temp.

Answer 1

pH of Ba(OH)₂:
Barium hydroxide octahydrate consist of barium hydroxide,and it is possible to find mass of last one from mass of first one:
5.0756 mg = 0.0050756 g
0.0050756
m
Ba(OH)₂*8H₂O -------> Ba(OH)₂
171+8*18 171
m = 0.0027553g
amount is n = m/M_w = 0.0027553/171 = 1.661*10^-5 mol (in 4.6 L)
M = amount in 1 L
M = 1.661*10^-5 / 4.6 = 3.5 * 10^-6 M
Ba(OH)₂ <=> 2OH^- + Ba²⁺
[OH^-]= 2 M = 7 * 10^-6
pOH = -log [OH-] = 5.155
pH = 14 - pOH = 14 - 5.155 = 8.845

pH of H₂SO₄:
n = m/M_w = 0.19789/98 = 0.00202 mol (in 374 ml)
M = 0.00202/0.374 = 0.0054 M
[H⁺] = 2 M = 0.0108 M
pH = -log [H⁺] = 1.97