A solution contains 0.364 M HA (Ka 2.23.10 and 0.751 M NaA. What is the pH of th
ID: 593602 • Letter: A
Question
A solution contains 0.364 M HA (Ka 2.23.10 and 0.751 M NaA. What is the pH of this solution? What is the pH of this solution after 0.143 mol of HCI are added to 1.00 L of this solution? What is the pH of this solution after 0.286 mol of HCI are added to 1.00 L of this solution? #9 Points possible: 1 . Total attempts: 0 How many mol of solid LiA would have to be added to 1.0 L of a 3.917 M HA solution in order to obtain a buffer of 3.05? (The Ka for HA is 6. 5-10-4. Preview mol LiA #10 Points possible: 1 . Total attempts:0 Calculate the pH at the equivalence point for the titration of 4.441 M B: (a weak base with pKb 5.89) with 4.441 M HCI.Explanation / Answer
pka = -logka
= -log(2.23*10^-5)
= 4.65
pH = pka + log[NaA]/[HA]
= 4.65+log(0.751/0.364)
= 4.96
after addition of 0.143 mol HCl
pH = pka + log[(NaA-HCl)]/[(HA+HCl)]
= 4.65+log((0.751-0.143)/(0.364+0.143))
= 4.73
0.286 mol HCl
pH = pka + log[(NaA-HCl)]/[(HA+HCl)]
= 4.65+log((0.751-0.286)/(0.364+0.286))
= 4.5
9) pka = -logka
= -log(6.15*10^-4)
= 3.21
pH = pka + log[LiA]/[HA]
3.05 = 3.21+log(x/3.917)
x = no of mol of solid LiA = 2.71 mol
10) at equivalence salt formation takes place
concentration of salt = 4.41/2 = 2.205 M
pH = 7-1/2(pkb+logC)
pH = 7-1/2(5.89+log2.205)
= 3.88
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