A 50mL solution of 0.2MCH3COOH is titrated with 0.2MNaOH. Find the pH of solution when: (a) 0mLNaOH is added (pHinitial) (b) 10mL of NaOH has been added (pH10mL)
Given: Ka(CH3COOH)=1.8×10−5
A
pHinitial=2.72,pH10mL=4.14
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B
pHinitial=5.7,pH10mL=6.72
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C
pHinitial=1.72,pH10mL=7.28
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D
pHinitial=4.57,pH10mL=2.17
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Solution
The correct option is ApHinitial=2.72,pH10mL=4.14 Case(a) mmol of CH3COOH=50×0.2=10 mmol of NaOH=0 CH3COOH(aq)+NaOH(aq)→CH3COONa(aq)+H2O(l)10000 Since only weak acid CH3COOH is present initially. pHinitial=−log√KaCpHinitial=−12(log(1.8×10−5×0.2))pHinitial=−12(log(3.6×10−6))pHinitial=12(6−0.56)=5.442pHinitial=2.72
Case (b) mmol of CH3COOH=50×0.2=10 mmol of NaOH=10×0.2=2
After titration we can see that 8mmol of CH3COOH and 2mmol of CH3COONa are present. Final Concentration =Moles Total Volume Total Volume =50+10=60mL [CH3COOH]=860M [CH3COONa]=260M An acidic buffer is formed. So, pH=pKa+log([CH3COONa][CH3COOH])pH=4.74+log⎛⎝260860⎞⎠pH=4.74+log(14)=4.74−0.6pH10mL=4.14