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In which of the following mixed aqueous solutions pH = pk_a at equilibrium? (1) 100 mL of 0.1 M CH_3 COOH + 100 mL of 0.1 M CH_3 COONa (2) 100 mL o f 0.1 M CH_3 COOH + 50 mL o f 0.1 M NaOH (3) 100 mL o f 0.1 M CH_3 COOH + 100 mL o f 0.1 M NaOH (4) 100 mL o f 0.1 M CH_3 COOH + 100 mL o f 0,1 M NH_3 |
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Answer» (1) is correct. ` (1) 100 m L` of 0.01`M CH_3COOH + 100 m L` of ` 0.1 M CH_3COONa` ` [CH_3 COOH]=(100 xx 0.1)` `[CH_3 COONA]=( 100 xx 0.1)` themixtureis anacidicbuffer usinghendersonequationwe get ` pH =pK_a + log([CH_3 COONa])/([CH_3 COOH])` ` impliesPH =pK_a+ og((100 xx 0.1 ))/( ( 100 xx 0.1 )) impliespH =pK_a+ log 1` `impliespH= pk_a( :.log1=0)` (2)100 mLof `0.1M CH)3 COOH+ 50mL ` of`0.1M NaOh `  WIDTH="80%"> Remaining 5milliole0 millimole5 millimole5millimole so anacidicbufferis forward ` thereforepH = pK_a+ log(["salt"])/(["acid"])` `impliespH=pK_a + log (5)/(5)IMPLIES PH= pK_a ` `(:.log (5)/(5) =- log1=0)` `(3)100m L` of ` 0.1 M CH_3COOH+100 mL` of ` 0.1 M` ` NaOH `  Now`CH_3COONa ` whichisa saltfromweakacidandstrongbasewillundergohydrolysis(anionichydrolysisandthe pHwillbedeter minedas below `pH =7 +1/2 pK_a+1/2log c` so` pHnepK_a ` (4) `100 mL` of `0.1M CH_3 COOH + 100mL`of 0.1 M `NH_3` `[CH_3] or [NH_4OH ] =( 100 xx 0.1) =10` ` (##MTG_WB_JEE_CHE_C12_E02_010_S03.png" width="80%"> Now`CH_3 COONa` whichis a saltfromweakacidand strongbasewill underogohydrolysis( anionichydrolysis) andthe pH will bedetereminedas below `pH= 7 + 1/2pK_a+1/2log c` So `PHnepK_a ` (4) 100 mLof 0.1 `CH_3COOH+ 100 mL ` of`0.1MCH_3` `[CH_3 COOH ]=(100 xx 0.1 )=10` `[NH_3 COOH ]=(100 xx 0.1 )=10`  `pH =7 +1/2(pk_a-pK_b)` so `pHnepK_a ` |
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