A. A buffer solution is 0.349 M in HNO 2 and 0.316 M in KNO 2 . If K a for HNO 2
ID: 881218 • Letter: A
Question
A. A buffer solution is 0.349 M in HNO2 and 0.316 M in KNO2. If Ka for HNO2 is 4.5E-4, what is the pH of this buffer solution?
B.A buffer solution is 0.310 M in H2CO3 and 0.261 M in NaHCO3. If Ka for H2CO3 is 4.2E-7, what is the pH of this buffer solution?
C.What is the pH at the equivalence point in the titration of a 15.0 mL sample of a 0.334 M aqueous acetic acid solution with a 0.467 M aqueous barium hydroxide solution? Ph=??
D.A 39.9 mL sample of a 0.438 M aqueous hypochlorous acid solution is titrated with a 0.346 M aqueous sodium hydroxide solution. What is the pH after 20.6 mL of base have been added?
pH =
Explanation / Answer
A- pH =- log Ka + log [ conjugate base/acid]
pH = -log4.5x10^-4 + log [0.316/0.349]
pH =3.3468 -0.0431 = 3.3037
B-pH = -logKa + log [conjugate base / acid]
pH = - log 4.2 x10^-7 + log [0.216/0.310]
pH = 7- 0.6232+ log 0.8419
pH = 6.3768 - 0.074739= 6.30206
C)2CH3COOH + Ba(OH)2 ---> (CH3COO)2Ba + 2H2O
mole of acid/2 = mole of base required at eq. point =0.334 x15ml/2 = 2.505mmol
volume of base required = 2.505x10^-3mol/.467 M = 5.364 ml
total volume = 15 + 5.364= 20.364ml
molarity of barium accetate = 2.505 x10^-3 / 20.364 x10^-3 = 0.123M
Kw = KaKb
1.00 x 10-14 = (1.77 x 10-5 ) (y)
y= 5.65 x 10-10
5.65 x10^-10 = x^2 /0.123
x= .8336x10^-5
pOH = 4.921
pH = 14-4.921 = 9.079
D-HClO + NaOH ---->NaClO + H2O
mole of acid = 39.9ml x 0.438 = 17.4762 x10^-3 mol
mole of base = 20.6ml x 0.346 = 7.1276mmol
amount of NaClO formed = 7.1276mmol
amount of acid left = 10.3486mmol
pH = pKa + log[salt/acid]
pH = 7.42 + log [7.1276/10.3486] = 7.42-0. 16197 = 7.258
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