1. Explain why you would not observe a precipitate when you mixed equal volumes

Question

1. Explain why you would not observe a precipitate when you mixed equal volumes of 0.01
BaCl2 is mixed with 0.1 Pb(NO3)2. The Ksp for PbCl2 is 2.4 x 10-4

2. Barbituric acid, HC4H3N2O3, a monoprotic acid, is used to prepare
various barbituate drugs. (a) Calculate the concentration of hydrogen
ion and pH in 100.0 mL of 0.250M solutoin of the acid. The value of Ka
is 9.8 x 10-5. (b) If 25.0 mL of 0.50M NaOH is added to the above
solution, what is the pH of the resluting solution now? (d) What is the
pH of the resulting solution after 55.0 mL of 0.50M NaOH?

Explanation / Answer

[Pb] = 0.1/2 = 0.05

[Cl2] = 0.01*2/2 = 0.01

k= 0.05*(0.01)^2 = 0.000005 > ksp

therefore precipitate is not observed

ka = X^2/ (0.25 - X) = 9.8*10^-5

X = 0.0049

pH = -log 0.0049 = 2.31

acid = (100*0.25-25*0.5)/125 = 0.1

salt = 25*0.5 / 125 = 0.1

pH = pka + log [salt]/[acid] = 4 + log(0.1/0.1) = 4

[OH] = (55*0.5 - 100*.25)/155 = 0.016129

p[OH] = 1.7924

p[H] = 14-1.7924 = 12.2076

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