Strong base is dissolved in 595 mL of 0.200 M weak acid (Ka 3.84 x109) to make a
ID: 562179 • Letter: S
Question
Strong base is dissolved in 595 mL of 0.200 M weak acid (Ka 3.84 x109) to make a buffer with a pH of 3.93. Assume that the volume remains constant when the base is added. HA(aq) +0H-(aq) H,O) + A-(aq) Calculate the pK, value of the acid and determine the number of moles of acid initially present Number Number When the reaction is complete, what is the concentration ratio of conjugate base to acid? Number How many moles of strong base were initially added? Number mol OH O Previcus (d) Check Answer 0 Next Exit HintExplanation / Answer
¡) HA(aq) <-------> H+ (aq) + A-(aq)
Ka = [H+][A-]/[HA] = 3.84×10^-5
Ka = 3.84×10^-5
pKa = -logKa
= -log(3.84×10^-5)
= 4.42
No of mole of acid = (0.200mol/1000ml)×595ml = 0.119mol
ii) Henderson- Hasselbalch equation is
pH = pKa + log([A-]/[HA])
3.93 = 4.42 + log([A-]/[HA])
log([A-]/[HA]) = -0.49
[A-]/[HA] = 1×10^-0.49
= 0.324
¡¡¡) [A-]/[HA] = 0.324
[A-] = 0.324[HA]
total concentration of buffer is 0.200M
[A-] + [ HA ] = 0.200M
0.324[HA] + [ HA ] = 0.200M
1.324[HA] = 0.200M
[ HA ] = 0.200M/1.324 = 0.1511M
[A-] = 0.200M - 0.1511M = 0.0489M
No of mole of A- = (0.0489mol/1000ml)×595ml = 0.0291mol
reaction between acid and base is
HA(aq) + OH-(aq) -------> H2O(l) + A-(aq)
So,
No of mole of A- = No of mole of strong base added
Therefore,
moles of strong base initially added = 0.0291