A laboratory manager wants to prepare 250.00 mL of a buffer with a pH of 4.763 from equimolar solutions of HN3 (Ka = 1.900e-5) and LiN3. What volume of the HN3 solution would be required?
Please....Help....spent way too much time on this....for nothing :(
Please....Help....spent way too much time on this....for nothing :(
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LiN3 --> Li+ + 3N 1/3-.....each N has a -1/3 charge
HN3 = weak acid dissociates to H+ and N 1/3-
Ka HN3 = 1.9x10^-5 so pKa = 4.72
for this buffer to have a pH of 4.763, the [N 1/3-]/[HN3] needs to be determined
pH = pKa + log[salt]/[acid]
4.763 = 4.72 + log[salt]/[acid]
0.043 = log[salt]/[acid]
10^0.043 = [salt]/[acid] = 1.1
so, the M of N 1/3- is 1.1 x greater than the M of HN3
1.1M x 0.25L = 0.276moles required of N 1/3- and HN3
xL of 1M = 0.276moles HN3
volume = 0.276L
to get 250ml total volume, 0.276L / 0.250L = 1.1
so, the volume of HN3 = 1.1> than the volume for LIN3
for every 1ml LiN3 added, 1.1ml HN3 is to be added
0.276 / 4 = 0.069L HN3 required 0.062L LiN3 required. bring up to 0.25L
HN3 = weak acid dissociates to H+ and N 1/3-
Ka HN3 = 1.9x10^-5 so pKa = 4.72
for this buffer to have a pH of 4.763, the [N 1/3-]/[HN3] needs to be determined
pH = pKa + log[salt]/[acid]
4.763 = 4.72 + log[salt]/[acid]
0.043 = log[salt]/[acid]
10^0.043 = [salt]/[acid] = 1.1
so, the M of N 1/3- is 1.1 x greater than the M of HN3
1.1M x 0.25L = 0.276moles required of N 1/3- and HN3
xL of 1M = 0.276moles HN3
volume = 0.276L
to get 250ml total volume, 0.276L / 0.250L = 1.1
so, the volume of HN3 = 1.1> than the volume for LIN3
for every 1ml LiN3 added, 1.1ml HN3 is to be added
0.276 / 4 = 0.069L HN3 required 0.062L LiN3 required. bring up to 0.25L