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Homework Statement
find the pH of:
100mL x .20M HOAc + 100mL x.10M Ba(OH)2
The Attempt at a Solution
100x.2 = 20mmol HOAc
100x.1 = 10mmol Ba(OH)2
20-10=10
10/200=.05M
pOH = -log(.05) = 1.3
pH=14-1.3=12.7
Is this correct?
hint, hint, the situation is going to be "neutralized" as chemisttree has saidHomework Statement
find the pH of:
100mL x .20M HOAc + 100mL x.10M Ba(OH)2
The Attempt at a Solution
100x.2 = 20mmol HOAc
100x.1 = 10mmol Ba(OH)2
20-10=10
10/200=.05M
pOH = -log(.05) = 1.3
pH=14-1.3=12.7
Is this correct?
That's what happens when English is not your first language... :grumpy:Ba(OH)2 is not diprotic. Dibasic.