Solutions available 1.Acetic acid (CHCOH, K = 1.8 10) and sodium acetate (NaCHO)
ID: 892579 • Letter: S
Question
Solutions available
1.Acetic acid (CHCOH, K = 1.8 10) and sodium acetate (NaCHO).
2.Ammonium chloride (NHCl, K for NH = 5.6 10) and ammonia (NH).
(The buffer will be prepared by choosing the appropriate acid-base pair, calculating the molar ratio of acid to base that will produce the assigned pH, and then mixing the calculated amounts of the two compounds with enough deionized water to make 200. mL of buffer solution.
A solution with approximately the same pH as the buffer solution will be prepared by diluting a solution of a strong acid or a strong base. This solution will not be a buffer solution, as can be shown by comparing its buffering ability to that of the buffer solution. Finally, a known amount of strong acid or base will be added to the buffer solution. Before adding this acid or base, the pH change that the addition should cause will be calculated. The observed pH change will be compared to the calculated value.)
If pH =5.45, Decide which of the two available buffer systems you should use.
3.Determine the ratio of [A]/[HA] needed in the buffer.
4.Which component will be more concentrated, A or HA? Choose a value for the concentration of that component anywhere in the range 0.2 – 0.6 M. Calculate what the concentration of the other component should be.
5.Calculate the volume (in mL) of 17.5 M CHCOH or 14.8 M NH needed to make 200 mL of your buffer. Calculate the mass (in g) of solid NaCHCO·3HO or NHCl needed to make 200 mL of your buffer.
Explanation / Answer
For Q2, If pH =5.45 you have to choose Acetic acid-Sodium acetate buffer. Among these two buffer choice, one is weak acid-Salt buffer and other one is weak base-salt buffer. Since, pH is 5.45 you will have to choose acid buffer.
Q3: pH = pKa + log ([Salt] / [Acid])
Ka = 1.8 x 10-5 so, pKa = - log Ka = - log 1.8 x 10-5 = 5 - 0.2553 = 4.7447
5.45 = 4.7447 + log ([Salt] / [Acid])
log ([Salt] / [Acid]) = 0.7053
[Salt] / [Acid] = 5.0734
Q4: Choose [Acid] = [HA] = 0.2 M
log ([Salt] / [Acid]) = 0.7053
log [Salt] - log [Acid] = 0.7053
log [Salt] - 0.699 = 0.7053
log [Salt] = 0.0063
[Salt] = [A-] = 1.0146 M
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