You may work this problem using the Henderson-Hasselbalch equation. Technically that uss concentration of base and acid but you may also use mols and the answer comes out the same. I use mols although concentration is supposed to be substituted.
millimols HAc = mL x M = 50 x 0.1 = 5
mmols NaAc = 100 x 0.1 = 10
mmols NaOH = 0.937 x 1 = 0.937
........HAc + OH^- ==> H2O + Ac^-
I.......5....0.........0.....10
add........0.973.................
C..-0.973..-0.973...........+.973
E...4.027....0............10.973
Substitute the E line into the H-H equation and solve for pH.
100 mL 0.1 M sodium acetate solution is mixed with 50 mL 0.1 M acetic acid. If now 0.937 mL of 1 M sodium hydroxide is added to the buffer what is the change in pH? pKa (acetic acid) = 4.76
3 answers
Umm... Im in something grade. I don't know this stuff.
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