1) Which of the curves seems to have the most vertical line around the equivalen
ID: 1022415 • Letter: 1
Question
1) Which of the curves seems to have the most vertical line around the equivalence point? Explain why. (If you suspect that your data are inaccurate, explain which one you think should have been most vertical and why.)
2) Which of the curves seems to have the most sloping line around its equivalence point? Explain why. (If you suspect that your data are inaccurate, explain which one you think should have been most vertical and why.)
3) State the pH at the equivalence point for each of your four curves.
4) Write the equilibrium equations for each of the four titrations at their equivalence points. You may simply write the net ionic equations.
5) Is the pH at the equivalence point neutral for all of them? Explain why or why not for each of the four.
6) Which of the four systems formed a buffer (at least initially)? Explain why.
WILL RATE
pH vs. Volume (drops) for Strong acid (H2S02) and Strong Base (NAOH pH vs. Volume (drops) of weakarid (vinegar) andstrong (NaOH) 12 10 10 10 0 2C 25 0 pH vs. Volume (Drops) of weak acid (vinegar) and weak Ease (NH3) CH vs.Volume (droos) of Strong acid (H2S04) and weak base (ammon a) 10 10 20 25 30 35 Vclurme (drops) of ammcnia Vclum (drops) of ammonieExplanation / Answer
1) The first curve (Strong acid vs strong base) have the most vertical line around the equivalence point as there is a sharpe change in pH value between strong acid (low pH) and Strong base (high pH).
2) The Third curve (weak acid vs weak base) have the most sloping line around the equivalence point as there is no sharpe change in pH value between weak acid (moderate pH, ~5/6/7) and weak base (moderate pH, ~7/8/9).
3,4,5)* For strong acid and strong base at equivalence point pH=7
*For weak acid and strong base at equivalence point pH controlled by the hydrolysis of salt. pH=7+1/2pKa+1/2log[salt]
*For strong acid and weak base at equivalence point pH also controlled by the hydrolysis of salt but equation is pH=1/2pKa-1/2log[salt]
*For weak acid and weak base at equivalence point, pH=1/2PKa1+1/2PKa2+1/2log[B]/[A]
6) weak acid and weak base system will form buffer.
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