A buffer is prepared by combining 0.24mol of HA, K_a = 1.3 times 10^-5, and 0.35
ID: 504123 • Letter: A
Question
A buffer is prepared by combining 0.24mol of HA, K_a = 1.3 times 10^-5, and 0.35 mol of NaA in a total volume of 415 mL at 25 degree C. a. Calculate the equilibrium pH of the solution. b. 200.0 mL of 1.5 M HCl solution is added to the buffer above, and the resulting solution is allowed to come to equilibrium. Calculate the equilibrium pH of the solution. Answer the following questions about reaction below: S^2- (aq) + NO^-_3 (aq) rightarrow NO_2 (g) + S_8 (s) a. Oxidation half reaction: b. Reduction half reaction: c. Balanced Net reaction in acidic media:Explanation / Answer
5)
a) moles of acid = 0.24
moles of salt = 0.35
total volume = 415 mL = 0.415 l
[acid] = 0.24 / 0.415 = 0.578 M
[salt] = 0.35 / 0.415 = 0.84 M
pH = pKa + log [salt] /[acid]
pKa = - log Ka = - log [1.3 x 10-5]
pKa = 4.89
pH = 4.89 + log [0.84] / [0.578]
pH = 5.05
b) after 1.5 x 0.2 = 0.3 moles HCl added
[HA] = 0.24 + 0.3 = 0.54 moles
[NaA] = 0.35 - 0.3 = 0.05 moles
total volume = 415 + 200 = 615 mL = 0.615 L
[HA] = 0.54 / 0.615 = 0.88 M
[NaA] = 0.05 / 0.615 = 0.081 M
pH = pKa + log [NaA] / [HA]
pH = 4.89 + log [0.081] / [0.88]
pH = 3.85
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