- #1
Soaring Crane
- 469
- 0
Hydrocyanic acid is a weak acid (Ka = 4.9 x 10-10). If 0.1300 moles of gaseous HCN are dissolved in 0.8700 liters of water. Determine the pH of the HCN solution formed.
HCN + H2O <-> H3O+ + CN-
K_a = [CN-][H3O+]/[HCN]
M HCN = 0.1300 mol HCN/0.87351338 L = 0.14882428 M HCN ??
For L solution: 0.1300 mol HCN*(27.026 g/1 mol HCN) = 3.51338 g HCN
870 g H2O + 3.51338 g HCN = 873.51338 = 873.51338 mL = 0.87351338 L??
K_a = [x][x]/[0.14882428 - x]
Assuming 0.14882428 - x = 0.14882428,
x = sqrt(4.9 x 10-10*0.14882428) = 8.539549E-6 = [H3O+]
pH = -log(8.539549E-6) = 5.068565065 = 5.07 ?
Thanks.
HCN + H2O <-> H3O+ + CN-
K_a = [CN-][H3O+]/[HCN]
M HCN = 0.1300 mol HCN/0.87351338 L = 0.14882428 M HCN ??
For L solution: 0.1300 mol HCN*(27.026 g/1 mol HCN) = 3.51338 g HCN
870 g H2O + 3.51338 g HCN = 873.51338 = 873.51338 mL = 0.87351338 L??
K_a = [x][x]/[0.14882428 - x]
Assuming 0.14882428 - x = 0.14882428,
x = sqrt(4.9 x 10-10*0.14882428) = 8.539549E-6 = [H3O+]
pH = -log(8.539549E-6) = 5.068565065 = 5.07 ?
Thanks.