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Question

Given two mixtures: (A) NaOH and Na2CO3 and (B) NaHCO3 and Na2CO3.
100 mL of mixture (A) required a mL and b mL of 1 M HCl in separate titration using phenolphthalein and methyl orange indicators respectively while 100 mL of mixture (B) required x mL and y mL of same HCl solution is separate titration using the same indicators.

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Solution

(A) Mixture A: (NaOH+Na2CO3)
pmEqqmEq–––––––––––
100 mL
With phenolphthalein: (NaOH reacts completely and 1/2mEq of Na2CO3 reacts)
p+q2=a×1 ....(i)
with methyl orange (both react completely):
p+q=b×1 ....(ii)
Solve: q2=(ba),q=2(bq)
mmoles of Na2CO3(q)=2(ba)
mmoles of Na2CO3(q)=2(bq)2=(ba)mmol
(B) Mixture B: (NaHCO3+Na2CO3)
rmEqsmEq–––––––––––
100 mL
With phenolphthalein: (NaHCO3 does not react)Na2CO3 react half)
s2=x×1mEq of Na2CO3
s=2xmEq of Na2CO3=2x2
=xmmol of Na2CO3
With methyl orange (both reacts completely):
r+s=y×1 mEq
r+2x=y
(D)r=(y-2x) mEq =mmolesofNaHCO_{3}(C)NaOH$ in Mixture A:
Solve equations (i) and (ii)
2p+32q=a+b
Substitute the value of q=2(ba)
2p+3/2×/2(ba)=a+b
2p=a+b3b+3a=4a2b
p=(2ab)MEq= mmoles of NaOH

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