Interactive titration Worksheet Name Buffers A buffer solution is one which main

Question

Interactive titration Worksheet Name Buffers A buffer solution is one which maintains an approximately constant pH when small amou a strong acid or base are added. Solutions containing either weak acids and their or weak bases and their conjugate acids, can be buffering solutions s bases 1. The number of moles of the conjugate pairs must be large compared to the moles strong acids or bases. 2. The ratio of the conjugate pair (weak acidyconjugate base] or (weak basel / should lie between 0.1 and 10, with optimal buffering at a 1:1 ratio. the conjugate pair (lweak acidý/conjugate base] or (weak base]/[conjugate acidl) 3. The pH of a 1:1 ratio buffer is equal to the pKe of the weak acid or pKo of the weak base. The effective range is t 1 from the pKa or pKb CHaCOOH/CHsCOO K 1.8x 10 NHs/NH4 K 1.8 x 10 K-5.6 x 10-11 K 6.2 x 10 ????./CO32. 1. Which of the conjugate pairs, shown above would be appropriate for preparing a buffer at pH- 7.0? 2. What is the pH of a buffer formed from 50 mL of 15.0 M NHs and 53.5 g of NH+CI in enough water to make 500 mL of solution? 3. What is the pH of the previous solution after addition of 100 mL of 0.2 M NaOH?

Explanation / Answer

Question 1.

The conjugate pair H2PO4-/HPO42- would be appropriate for preparing a buffer at pH = 7.0

Explanation: pKa of H2PO4- = -Log(Ka)

= -Log(6.2*10-8)

= 7.2

Here, pKa is close to given pH and hence the conjugate pair H2PO4-/HPO42- would be appropriate for preparing a buffer.

Question 2.

The no. of moles of NH3 (nNH3) = 50*10-3 L * 15 mol/L = 0.75 mol

The no. of moles of NH4Cl (nNH4Cl) = 53.5 g/(53.5 g/mol) = 1 mol

According to Henderson-Hasselbulch equation:

pH = pKa + Log(nNH3/nNH4Cl)

pKb of NH3 = -LogKb = -Log(1.8*10-5) = 4.74

i.e. pKa = 14 - pKb = 14-4.74 = 9.26

i.e. pH = 9.26 + Log(0.75/1) = 9.14

Question 3.

The no. of moles of NaOH (nNaOH)= 100*10-3 L * 0.2 mol/L = 0.02 mol

Now, pH = pKa + Log(nNH3+?nNaOH/nNH4Cl-?nNaOH)

i.e. pH = 9.26 + Log(0.75 + 0.02/1-0.02) = 9.16

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