On the basis of the information above, a buffer with a pH = 9 can best be made b
ID: 1065616 • Letter: O
Question
On the basis of the information above, a buffer with a pH = 9 can best be made by using pure NaH_2PO_4. H_2PO_4^- + PO_4^3-. H_2PO_4^- + HPO_4^2-. HPO_4^2- +PO_4^3- None of these choices would be suitable for preparing a pH 9 buffer. 50.00 mL of 0.10 M HNO_2 (nitrous acid, K_a = 4.5 times 10^-4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH solution is added, the pH in the titration flask will be 2.17 3.35 2.41 1.48 7.00 What mass of sodium fluoride must be added to 250. mL of a 0.100 M HF solution to give a buffer solution having a pH of 3.50? (K_a(HF) = 7.1 times 10^-4) 0.49g 1.5g 3.4g 2.3g 75gExplanation / Answer
A) Ka of H2 PO4- is 8 x 10-8
Thus to prepare a buffer of pH = 9 the closer pKa is that of H2PO4 - ion only.
pH of buffer = pKa + log [salt]/ [acid]
thus the best possible combinationamong the givenspecies to get buffer of pH = 9 is H2PO4- + HPO4-2
That is OPTION C
Q4) HA + KOH ------> KA + H2O
50x 0.10 - 0 0 initial mmoles
25x0.10 change
2.5 0 2.5 - equilibrium mmoles
The solution forms a buffer whose pH is given by Hendersen equation as
pH = pKa + log [conjugate base]/[acid]
= (4- log 4.5) + log (2.5/75)/(2.5/75)
= 3.347
Q5) using the same equation
3.5 = (4 - log 7.1 ) + log x/(250x0.1)
solving for x , we get x = mmoles of NaF required = 56.125 mmoles
mass of NaF = (mmoles x 10-3) xmolar mass
= 56.125 x 10-3 x42 g
= 2.3575 g
thus option d is correct.
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