Asked by Mikey

a buffer is prepared by mixing .208 L of 0.452 M HCl and 0.5 L of 0.4 M sodium acetate. When the Ka= 1.8 x 10^-5

I figured out that the pH= 4.74

but then it asks how many grams of KOH must be added to the buffer to change the pH by .155 units?
Answered by DrBob222
Show me how you obtained pH = 4.74. I can't get that. My answer looked more like 4.79.
Answered by Mikey
Since Ka a measure of H+ ions,
-log(Ka)= pH.. my assignment says that part is right, I just don't know what to do from there to get M -> moles -> grams
Answered by DrBob222
-log Ka = pKa which may or may not be = pH. Ka is a measure of how strong/weak the acid is.