ANSWER :
pKa = 4.518
No. of moles of weak HA present initially = 0.381 mol
[A - ] / [HA] = 0.365
moles of strong base initially added to acid solution = 0.102 mol
1) Calculation of pKa
We have, pKa = -log Ka = - log ( 3.03 10 -05 ) =4.518
2) Calculation of no. of moles of acid HA present initially.
We have, Molarity = No. of moles of solute / Volume of solution in L
On rearranging, we get No. of moles of solute = Molarity volume of solution in L
No. of moles of weak acid HA = 0.600 mol / L 0.635 L = 0.381 mol
ANSWER: No. of moles of weak HA present initially = 0.381 mol
3)
We have Henderson's equation for acidic buffer pH = pKa + log [Salt] / [ Acid]
4.08 = 4.518 + log [A - ] / [HA]
log [A - ] / [HA] = 4.08 - 4.518 = -0.438
[A - ] / [HA] = 10 -0.438 =0.365
ANSWER : [A - ] / [HA] = 0.365
4)
Consider a given reaction , HA + OH - A - + H2O
Let's use ICE table
Concentration (moles) | HA | OH - | A - |
Initial | 0.381 | X | |
Change | - X | -X | +X |
Equilibrium | 0.381 -X | 0.0 | X |
Therefore, [A - ] / [HA] = X / 0.381 - X
We have calculated [A - ] / [HA] = 0.365
Hence, X / 0.381 - X = 0.365
X = 0.365 ( 0.381 -X) = 0.139 - 0.365 X
X + 0.365 X = 0.139
1.365 X = 0.139
X = 0.139 / 1.365
X = 0.102
Hence, moles of strong base initially added to acid solution = 0.102 mol
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