Question

#CHEMISTRY

Q. 50 ml of 0.1 M NaOH is added to 75 ml of 0.1 M $NH_{4}CI$ to make a basic buffer. If $pK_{a}$ of ${NH_{4} }^{ + }$ is 9.26, then what is the pH of the solution?

ANSWER THE QUESTION WITH PROPER EXPLANATION

#No Spamming Plz​

Answer 1

Solution :

⏭ Given:

✏ Conc. of NaOH = 0.1M

✏ Volume of NaOH = 50ml

✏ Conc. of NH4Cl = 0.1M

✏ Volume of NH4Cl = 75ml

✏ pKa of NH$_4$$^+$ = 9.26

⏭ To Find:

✏ The pH of the solution

⏭ Formula:

✏ Formula of pOH for basic buffer solution is given by...

$\bigstar \: \boxed{ \sf{ \pink{ pOH = pKb + log_{10}\frac{[salt]}{[base]} }}} \: \bigstar$

⏭ Calculation:

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• As per concept of molarity

✒ Initial moles of NaOH = 0.1×50/1000 = 0.005 moles

✒ Initial moles of NH4Cl = 0.1×75/1000 = 0.0075 moles

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• At t = 0 times

NH4Cl + NaOH ==>> NaCl + NH4OH

(0.0075) (0.005) (0) (0)

• At t = t times

NH4Cl + NaOH ==>> NaCl + NH4OH

(0.0025) (0) (0.005) (0.005)

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• Basic buffer (NH4OH + NH4Cl)

✏ NH4OH = weak base

✏ NH4Cl = salt of (weak base + strong acid)

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• Calculation of concentration

✏ Total volume of solution = 125ml

✒ Conc. of salt (NH4Cl) = 0.0025×1000/125 = 0.02 M

✒ Conc. of weak base (NH4OH) = 0.005×1000/125 = 0.04 M

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• Calculation of pKb

✏ pKa + pKb = 14

✏ 9.26 + pKb = 14

✏ pKb = 4.74

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• Calculation of pH

✒ pOH = pKb + log$_10$ [salt]/[base]

✒ pOH = 4.74 + log$_10$ 0.02/0.04

✒ pOH = 4.74 + log$_10$ (0.5)

✒ pOH = 4.74 - 0.3

✒ pOH = 4.44

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✏ pH + pOH = 14

✏ pH + 4.44 = 14

✏ pH = 14 - 4.44

✏ pH = 9.56

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$\therefore$ pH of basic buffer solution = 9.56