How do you solve this problem?
What is the pH of a solution formed by dissolving 12g of Ba(OH)2 in 500mL of water?
Thank you!
What is the pH of a solution formed by dissolving 12g of Ba(OH)2 in 500mL of water?
Thank you!
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Ba(OH)2 <==> Ba^2+ + 2OH^-
12g * (1mol/171.34g) = 0.07 mol Ba(OH)2
0.07 mol / 0.5L = 0.14 M of Ba(OH)2
There is a 1:2 ratio between Ba(OH)2 and the hydroxide ion so the concentration of OH is 2 * 0.14M = 0.28 M of OH
[H][OH] = 1 * 10^-14
[H][0.28M] = 1 * 10^-14
[H] = 3.57 * 10^-14 M
pH = -log[H]
pH = -log[3.57 * 10^-14 M]
pH = 13.45
12g * (1mol/171.34g) = 0.07 mol Ba(OH)2
0.07 mol / 0.5L = 0.14 M of Ba(OH)2
There is a 1:2 ratio between Ba(OH)2 and the hydroxide ion so the concentration of OH is 2 * 0.14M = 0.28 M of OH
[H][OH] = 1 * 10^-14
[H][0.28M] = 1 * 10^-14
[H] = 3.57 * 10^-14 M
pH = -log[H]
pH = -log[3.57 * 10^-14 M]
pH = 13.45