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Questions  >>  JEEMAIN and NEET  >>  Chemistry  >>  Equilibrium
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Q)

$pH$ of the solution containing 50.0 mL of 0.3 M HCl and 50.0 mL of 0.4 M $NH_3$ is

[Given $pK_a (NH^+_4)$ = 9.26]

 

(a) 4.74

(b) 9.26

(c) 8.78

(d) 4.63

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A)
Toolbox:
  • pOH=pKb+log[Salt][Base]" role="presentation" style="position: relative;">pOH=pKb+log[Salt][Base]pOH=pKb+log[Salt][Base]pOH = pK_b + log \frac{[Salt]}{[Base]}
 
Answer: 8.78
 
pKa(NH4+)=9.26" role="presentation" style="position: relative;">pKa(NH+4)=9.26pKa(NH4+)=9.26pK_a(NH^+_4) = 9.26
 
∴pKb(NH3)=pKw−pKa(NH4+)" role="presentation" style="position: relative;">pKb(NH3)=pKwpKa(NH+4)∴pKb(NH3)=pKw−pKa(NH4+)\therefore pK_b (NH_3) = pK_w - pK_a(NH^+_4) = 14 - 9.26 = 4.74
 
On adding 15 m mol (50 mL of 0.3 M) HCl to 20 m mol (50 mL of 0.4 M) NH3" role="presentation" style="position: relative;">NH3NH3NH_3 the resulting solution contains 15 m mol (in 100 mL) of NH4Cl" role="presentation" style="position: relative;">NH4ClNH4ClNH_4Cl and 5 m mol (in 100 mL) of NH3" role="presentation" style="position: relative;">NH3NH3NH_3.
∴[Salt]=15×10−2M,[Base]=5×10−2M" role="presentation" style="position: relative;">[Salt]=15×102M,[Base]=5×102M∴[Salt]=15×10−2M,[Base]=5×10−2M\therefore [Salt] = 15 \times 10^{-2} M , [Base] = 5 \times 10^{-2} M
 
pOH=pKb+log[Salt][Base]=4.74+log15×10−25×10−2=4.74+log3=4,74+0.48=5.22" role="presentation" style="position: relative;">pOH=pKb+log[Salt][Base]=4.74+log15×1025×102=4.74+log3=4,74+0.48=5.22pOH=pKb+log[Salt][Base]=4.74+log15×10−25×10−2=4.74+log3=4,74+0.48=5.22pOH = pK_b + log \frac{[Salt]}{[Base]} = 4.74 + log \frac{15 \times 10^{-2}}{5 \times 10^{-2}} = 4.74 + log 3 = 4,74 + 0.48 = 5.22
 
pH = 14 - pOH = 14 - 5.22 = 8.78
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