- #1
Fibonacci
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am really stuck on this problem, can't seem to get to the right answer
Answer. 13.96 mL
pKa + log([base]/[acid])=pH
HCO3(-) --> H+ + CO3 (2-)
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
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