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1. A mixture of acids is made that has 0.0500M HNO3 and 0.0300M HC2H3O2. The Ka

ID: 227929 • Letter: 1

Question



1. A mixture of acids is made that has 0.0500M HNO3 and 0.0300M HC2H3O2. The Ka for HC2H3O2 is 1.8x10^-5. Solve for the pH of the mixture of acids.
2. Determine the pH of a solution that is 0.035M C2H8N2 where Kb(C2H8N2) = 8.3x10^-5?
3. Which of the following is a weak base? a. CH3NH2 b.LiOH c.HNO3 d.KCl e. KOH
4. What is the pH of a 0.100M solution of NaC2H3O2 when Ka(HC2H3O2)= 1.8x10^-5?
5. Classify the compound K2CO3 as acidic, basic or neutral. The Ka(HCO3-) is 5.6x10^-11?

1. A mixture of acids is made that has 0.0500M HNO3 and 0.0300M HC2H3O2. The Ka for HC2H3O2 is 1.8x10^-5. Solve for the pH of the mixture of acids.
2. Determine the pH of a solution that is 0.035M C2H8N2 where Kb(C2H8N2) = 8.3x10^-5?
3. Which of the following is a weak base? a. CH3NH2 b.LiOH c.HNO3 d.KCl e. KOH
4. What is the pH of a 0.100M solution of NaC2H3O2 when Ka(HC2H3O2)= 1.8x10^-5?
5. Classify the compound K2CO3 as acidic, basic or neutral. The Ka(HCO3-) is 5.6x10^-11?

1. A mixture of acids is made that has 0.0500M HNO3 and 0.0300M HC2H3O2. The Ka for HC2H3O2 is 1.8x10^-5. Solve for the pH of the mixture of acids.
2. Determine the pH of a solution that is 0.035M C2H8N2 where Kb(C2H8N2) = 8.3x10^-5?
3. Which of the following is a weak base? a. CH3NH2 b.LiOH c.HNO3 d.KCl e. KOH
4. What is the pH of a 0.100M solution of NaC2H3O2 when Ka(HC2H3O2)= 1.8x10^-5?
5. Classify the compound K2CO3 as acidic, basic or neutral. The Ka(HCO3-) is 5.6x10^-11?

Explanation / Answer

Ans 1

Balanced equation of HNO3

HNO3 = H+ + NO3-

I 0.05 0.05

Balanced equation of acetic acid with ICE table

HC2H3O2 = H+ + C2H3O2-

I 0.030 0.05

C - x +x +x

E 0.03-x 0.05+x x

Equilibrium constant expression of the reaction

Ka = [H+] [C2H3O2-] /[HC2H3O2]

1.8 x 10^-5 = x*(0.05+x)/(0.03-x)

Since 0.05 >> x so 0.05 - x = 0.05

And 0.03 >> x so 0.03 - x = 0.03

1.8 x 10^-5 = x*(0.05)/(0.03)

x = 0.0000108 = 1.08*10^-5

[H+] = 1.08*10^-5 M

pH = - log [H+] = - log (1.08*10^-5) = 4.96