How to Calculate Buffer Volume for Given pH and Concentrations

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To create a buffer with a pH of 10.30 using a 3.00 M HCO3- solution and a 0.200 M CO32- solution, the required volume of HCO3- is 13.96 mL. The calculation involves using the Henderson-Hasselbalch equation, where the pKa value is provided as 5.6e-11. An initial attempt yielded 11.93 mL, but this was incorrect. The discrepancy suggests a potential error in the initial calculations or assumptions. Clarification on the correct approach is sought to resolve the confusion.
Fibonacci
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am really stuck on this problem, can't seem to get to the right answer

Homework Statement

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

Homework Equations



pKa + log([base]/[acid])=pH

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

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

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