What is the pH of a 0.6 M solution of a weak base B with Kb = 1.41×10-5?

Question

What is the pH of a 0.6 M solution of a weak base B with Kb = 1.41×10-5?

A 25.0 mL sample of the base is titrated against 0.75 M HCl. What is the pH after 5.0 mL of the acid has been added?

What is the pH after 10.0 mL of acid has been added?

What is the pH when 30.0 mL of acid have been added?

What is the pH at the equivalence point?

What volume of base must be added to reach the equivalence point?

Answer 1

(a) The reaction:
BOH -> B+ + OH-
Let x= M of B+ and OH- formed from dissociation of BOH. Then
x^2/ [BOH] = 1.41x10-5
As an appx, ignore x compared to 0.6 M of BOH
x^2 = 8.5x10-6
x = 2.9 x 10-3 appx
We can find pOH and then pH
pOH = - Log x = - [Log10-3 + Log 2.9]
= 3 - 0.46 = 2.54 appx
pH = 14 - pOH = 11.46 appx

PART 2: The reaction is BOH + H+ -> B+ + H2O
The equilibrium relation that applies here is: [B-]/[BOH][H+] = Kb/Kw. You get this from the Kb expression above, mult num & denom by [H+] and realize that [H+][OH-]=Kw.
In the reaction, we have 0.025(0.6)= 0.015 mole of BOH in the sample to be titrated. 5 mL of acid supplies 0.005x0.75M = 0.00375 mole of H+. Virtually all H+ reacts, forming 0.00375 moles of B- in 30 mL of solution, which is 0.125 Molar. BOH moles are decreased from 0.015 to 0.01125 moles, and in 30 mL, this is 0.375 M.
Then we have
0.125 M/(0.375M)[H+]= 1.41x10-5/1x10-14
1/3[H+] = 1.41x10+9
Then 1/(4.2x10+9) = [H+] = 2.4x10-10
pH= - Log [H+]= 10 - Log 2.4 = 9.6 appx
REST OF PROBS. You can repeat the above analysis for the 10 mL acid addition.
With 30 mL acid, you have added 0.03x0.75 = 0.0225 moles of acid against 0.015 moles of base, so the situation is the same as if you added 0.0075 moles of acid in 55 mL of solution. Work back to find the [H+] and pH.

Answer 2

25ml of 0.7 M vulnerable base (Bec947a7cc943c84b1ef84958a7df827cec947a7cc943c84b1ef84958a7df827c) ability 25 x 0.7 = 17.5 meq of vulnerable base. then, further, 5ml of 0.55M ec947a7cc943c84b1ef84958a7df827cCl ability 2.seventy 5 meq of robust acid. ec947a7cc943c84b1ef84958a7df827cCl + Bec947a7cc943c84b1ef84958a7df827cec947a7cc943c84b1ef84958a7df827c ----------> BCl + ec947a7cc943c84b1ef84958a7df827c2ec947a7cc943c84b1ef84958a7df827c be conscious that BCl is a salt of robust acid and vulnerable base, so this is going to undergo cationic hydrolysis. so pec947a7cc943c84b1ef84958a7df827c = 7 - a million/2 { pKb + log c } as form of equivalents of acid and base in a neutralisation are equivalent, so 2.seventy 5 meq of BCl is shaped (as ec947a7cc943c84b1ef84958a7df827cCl is proscribing reagent) so taking some approximatiions like 5.01 = 5 and log11 = log 10, after putting values interior the formula- ph= 7- a million/2 { -log (5 x 10-8) + log(2.seventy 5)} pec947a7cc943c84b1ef84958a7df827c = 3.21055 (approx) be conscious that right here, we dont take contribution of ec947a7cc943c84b1ef84958a7df827cec947a7cc943c84b1ef84958a7df827c- ions from leftover Bec947a7cc943c84b1ef84958a7df827cec947a7cc943c84b1ef84958a7df827c beacuse Bec947a7cc943c84b1ef84958a7df827cec947a7cc943c84b1ef84958a7df827c is a vulnerable base and its dissociation willl added be deppressed by employing easy ion result from BCl.ec947a7cc943c84b1ef84958a7df827cence [ec947a7cc943c84b1ef84958a7df827cec947a7cc943c84b1ef84958a7df827c] is negligible.