Two buffer solutions are prepared: BUFFER A: 25mL of .100M NaOH are added to 30m
ID: 678790 • Letter: T
Question
Two buffer solutions are prepared: BUFFER A: 25mL of .100M NaOH are added to 30mL of .200MCH3COOH (acetic acid pKa=4.76) BUFFER B: 20 mL of .0100M NaOH are added to 30mL of .0200MCH3COOH (acetic acid pKa=4.76) a) What is the pH of buffer A? What is the pH ofbuffer B? b) If we add 1.00mL of 0.1M HCl to buffer A, what will theresulting pH be? c)If we add 1.00mL of 0.1M HCl to buffer B, what will theresulting pH be? d) briefly comment on the difference Two buffer solutions are prepared: BUFFER A: 25mL of .100M NaOH are added to 30mL of .200MCH3COOH (acetic acid pKa=4.76) BUFFER B: 20 mL of .0100M NaOH are added to 30mL of .0200MCH3COOH (acetic acid pKa=4.76) a) What is the pH of buffer A? What is the pH ofbuffer B? b) If we add 1.00mL of 0.1M HCl to buffer A, what will theresulting pH be? c)If we add 1.00mL of 0.1M HCl to buffer B, what will theresulting pH be? d) briefly comment on the differenceExplanation / Answer
Formula : pH = pKa + log ( No.of mols salt / No.of mols of acid ) a) CH3COOH + NaOH <---------- >CH3COONa + H2O Beforerxn 0.006 0.0025 0 Afterrxn 0.006 - 0.0025 0 0.0025 = 0.0035 pH = 4.76 + log ( 0.0025 / 0.0035 ) = 4.61 b) CH3COOH + NaOH <---------- > CH3COONa +H2O Beforerxn 0.0006 0.0002 0 Afterrxn 0.0004 0 0.0002 pH = 4.76 + log ( 0.0002 / 0.0004 ) = 4.45 c ) 0.0001 mols of acid is added to the acidic buffer no.ofmols of salt decreases and no.of mols of acid increases . No.of mols of salt =0.0025 mols - 0.0001 mols = 0.0024 mols No.of mols of acid = 0.0035mols + 0.0001 mols = 0.0036 mols pH = 4.76 + log ( 0.0024 / 0.0036 ) = 4.58 d ) 0.0001 mols of acid is added to the acidic buffer no.ofmols of salt decreases and no.of mols of acid increases . No.of mols ofsalt = 0.0002 mols - 0.0001 mols = 0.0001 mols No.of mols of acid = 0.0004mols + 0.0001 mols = 0.0005 mols pH = 4.76 + log ( 0.0001 / 0.0005) = 4.06 = 4.58 b) CH3COOH + NaOH <---------- > CH3COONa +H2O Beforerxn 0.0006 0.0002 0 Afterrxn 0.0004 0 0.0002 pH = 4.76 + log ( 0.0002 / 0.0004 ) = 4.45 c ) 0.0001 mols of acid is added to the acidic buffer no.ofmols of salt decreases and no.of mols of acid increases . No.of mols of salt =0.0025 mols - 0.0001 mols = 0.0024 mols No.of mols of acid = 0.0035mols + 0.0001 mols = 0.0036 mols pH = 4.76 + log ( 0.0024 / 0.0036 ) = 4.58 d ) 0.0001 mols of acid is added to the acidic buffer no.ofmols of salt decreases and no.of mols of acid increases . No.of mols ofsalt = 0.0002 mols - 0.0001 mols = 0.0001 mols No.of mols of acid = 0.0004mols + 0.0001 mols = 0.0005 mols pH = 4.76 + log ( 0.0001 / 0.0005) = 4.06 = 4.58 c ) 0.0001 mols of acid is added to the acidic buffer no.ofmols of salt decreases and no.of mols of acid increases . No.of mols of salt =0.0025 mols - 0.0001 mols = 0.0024 mols No.of mols of acid = 0.0035mols + 0.0001 mols = 0.0036 mols pH = 4.76 + log ( 0.0024 / 0.0036 ) = 4.58 d ) 0.0001 mols of acid is added to the acidic buffer no.ofmols of salt decreases and no.of mols of acid increases . No.of mols ofsalt = 0.0002 mols - 0.0001 mols = 0.0001 mols No.of mols of acid = 0.0004mols + 0.0001 mols = 0.0005 mols pH = 4.76 + log ( 0.0001 / 0.0005) = 4.06 = 4.58 = 0.0001 mols No.of mols of acid = 0.0004mols + 0.0001 mols = 0.0005 mols pH = 4.76 + log ( 0.0001 / 0.0005) = 4.06 = 4.58Related Questions
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