What is the pH of a sodium acetate solution prepared by adding 0.820 grams of sodium acetate to 100.0 ml of water at 25.0°C? The Ka at 25.0°C for acetic acid is 1.8 × 10-5.

the answer is 8.87

i dont know how? somehow u get
NaAc + h20 ----> HAc + NaOH

0.1 0 0

-x +x +x

0.1-x x x


how is there a 0.1 from this info?

Molarity of the sodium acetate solution is:

Molarity = (0.82g/82.0343g)*10



Kb = (1 * 10^-14) / Ka = 5.556* 10^-10

Kb = [x][x] / (Molarity - x)

Solve for [x] = 7.452 * 10^-6

pOH = -log[x] = 5.128

pOH + pH = 14

pH = 8.872


Edit: A fairly common practice is to just ignore the -x in the denominator of the bolded part, since as we know it is going to be very small compared to the Molarity so it isn't going to have a real impact on the final result and the equation is easier to solve without it there.

This post was edited by RzChaos on Oct 26 2015 04:36pm

thanks why is the molarity multiplied by 10 at the first step?

NaAc + H2O -> HAc + Na + OH is just the chemical reaction for this problem. Hopefully you understand what the deal there is, we are just adding sodium acetate to water and it dissociates into sodium and acetate ions. The acetate ion then accepts a proton from water and leaves a free hydroxide (OH) ion, making the resulting solution basic (i.e. with a Ph of more than 7). You wouldn't actually get NaOH like in your formula, sodium hydroxide completely dissociates in water.

You don't really need an ICE table for this like you did, because you aren't mixing two solutions together, but sure.

Initial NaAc concentration is 0.82 grams / 82.0343 g/mol / 0.1 L of water = 0.09996 g/mol, we can round to 0.1 g/mol, that's where the 0.1 came from. 82.0343 is the molar mass of sodium acetate.

It's the same as dividing by 0.1, which is the volume of water it's dissolved in.