# Homework Help: Acid Equilibrium

1. Mar 6, 2006

### Soaring Crane

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.

2. Mar 6, 2006

### GCT

set up seems fine to me

3. Mar 7, 2006

### Staff: Mentor

You have only two significant digits in the Ka, thus you may safely assume concentration is 0.13/0.87M - and neglect volume change.

Rest is OK.

4. Mar 8, 2006

### Haxx0rm4ster

That doesnt make a big difference, but it is not right. Just assume total V is .870 L .