Calculate the pH of buffer solution composed of 0.1 M weak base BOH and 0.2 M of its salt BA. [Kb = 1.8× 10⁻⁵

Problems · Problem 3.10

K_b = 1.8\times 10^{-5}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★

13% · 10/79 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

p $K_b = 4.7447$ and $\log(0.2/0.1) = 0.3010$ give pOH $= 5.0457$ and pH $= 8.9543$ .

Step 1. BOH (weak base) with its salt BA is a basic buffer, so the Henderson-Hasselbalch equation in its basic form (Eq. 3.26) applies: $\text{pOH} = \text{p}K_b + \log_{10}\dfrac{[\text{salt}]}{[\text{base}]}$ , with $[\text{salt}] = 0.2$ M and $[\text{base}] = 0.1$ M.

Step 2. $\text{p}K_b = -\log_{10}(1.8\times10^{-5}) = 5 - \log_{10}1.8 = 5 - 0.2553 = 4.7447$ .

Step 3. $\log_{10}\dfrac{0.2}{0.1} = \log_{10}2 = 0.3010$ , so $\text{pOH} = 4.7447 + 0.3010 = 5.0457$ . …

Start your 14-day free trial to unlock the full solution →

←Previous question · 3.8.4Problem 3.9 — Calculate the pH of buffer solution containing 0.05 mol NaF per litre …→Next section · 3.9Solubility product

Q.Calculate the pH of buffer solution composed of 0.1 M weak base BOH and 0.2 M of its salt BA. [ $K_b = 1.8\times 10^{-5}$ for the weak base]

Concept understanding — Buffer Solutions

Q.Calculate the pH of buffer solution composed of 0.1 M weak base BOH and 0.2 M of its salt BA. [Kb=1.8×10−5K_b = 1.8\times 10^{-5}Kb​=1.8×10−5 for the weak base]

🎓Unlock everything free for 14 days

Q.Calculate the pH of buffer solution composed of 0.1 M weak base BOH and 0.2 M of its salt BA. [ $K_b = 1.8\times 10^{-5}$ for the weak base]

Unlock everything free for 14 days