Buffer and pH

[Fibonacci]

am really stuck on this problem, can't seem to get to the right answer

1. The problem statement, all variables and given/known data


How many mL of a 3.00 M HCO3- solution must be added to 200 mL of a 0.200 M CO32- solution to make a buffer with pH = 10.30?

Answer. 13.96 mL

2. Relevant equations

pKa + log(/[acid])=pH

HCO3(-) --> H+ + CO3 (2-)



[b]3. The attempt at a solution

pKa + log([base]/[acid])=pH
pH=10.3
pKa= 5.6e-11 (this value is give to us in our book)

0.048188 = log([base]/[acid])
1.117=[base/acid]

since the final volume cancels
we simply just use the mole ratios
let x be the mL of HCO3 - needed

base= (0.20)(0.2)
acid= 3x

1.117=0.2(0.2)/3x

x=11.933ml, which is way different to the actual answer



would appreciated heaps if someone could help
thanks

[Borek]

Must be error in the answer (or some typo in the data), 11.93mL seems to be correct.

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