Asked by michelle
Calculate the pH of the solutions obtained by dissolving the following in a liter of water: 0.2 mole of glycine plus 0.1 mole of NaOH
I know the answer is ph = 9.60 which i know is pka2 of glycine but im not sure how to set up this question. This is just a practice problem for my homework which will have similar questions but i would like to know how to do my homework by knowing how to do this question.
Thank you
I know the answer is ph = 9.60 which i know is pka2 of glycine but im not sure how to set up this question. This is just a practice problem for my homework which will have similar questions but i would like to know how to do my homework by knowing how to do this question.
Thank you
Answers
Answered by
DrBob222
..NH2CH2COOH + NaOH =>NH2CH2CONa + H2O
I...0.2.........0..........0
add............0.1.............
C...-0.1.......-0.1........+0.1
E....0.1.........0.........0.1
pH = pKa glycine + log (base)/(acid)
base = salt = 0.1
acid = glycine = 0.1
pH = pKa + log 0.1/0.1
0.1/0.1 = 1 and log 1 = 0
so pH = pKa
I...0.2.........0..........0
add............0.1.............
C...-0.1.......-0.1........+0.1
E....0.1.........0.........0.1
pH = pKa glycine + log (base)/(acid)
base = salt = 0.1
acid = glycine = 0.1
pH = pKa + log 0.1/0.1
0.1/0.1 = 1 and log 1 = 0
so pH = pKa
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