1. Oct 15, 2007

### lananh

1. The problem statement, all variables and given/known data
Im having issues with two parts of this problem, not sure what I'm doing wrong...

A lake with a surface area of 12.2 acres (1 acre = 4.840 103 yd2) receives 1.67 in. of rain of pH 4.82. (Assume that the acidity of the rain is due to a strong, monoprotic acid.) How many moles of H3O+ are in the rain falling on the lake? I solved that the mols are 32 H3O.

If the lake is unbuffered (assume pH = 7.00) and its average depth is 9.20 ft before the rain, calculate the pH after the rain has been thoroughly mixed with lake water. (Ignore runoff from the surrounding land.)

New pH = IM HAVING ISSUES WITH THIS ONE

Natural waters are typically buffered by dissolved carbonates. If the lake originally contains 0.96 mM HCO3- and 0.28 mM H2CO3, what is the pH of the lake water before and after the rain? (Ignore runoff).

pH = (before rain) I solved it for 6.90
pH = (after rain) IM HAVING ISSUES WITH THIS ONE

any help or guidance would be greatly appreciated.

2. Relevant equations
ph=pka + log (a/ha)
ph=-log[h]

3. The attempt at a solution
above

2. Oct 17, 2007

### Gokul43201

Staff Emeritus
You've calculated the moles of H3O+ in the given volume of rainwater from the pH of rainwater. Next, you just need to work backwards. You know the number of moles of H3O+ from rain. Add to this the number of moles of H3O+ in the lake (pH=7), to get the total number of moles of H3O+. With this number and the final volume of the lake, you can calculate the final (unbuffered) pH.