EXAM 11-FI3 PRACTICE PROBLEMS The following values will Ixe useful for problems

Question

EXAM 11-FI3 PRACTICE PROBLEMS The following values will Ixe useful for problems in this chapses 4c HF HNO CH COOH HOCI HOBr K, 7.2 × 10- K,-4.5 10" K-1.8 x 10 K-3.5x 10* K,-2.5 × 104 NH CHibN -18 10 HCN HSo, H:CO K.-3.5 × 104 K,-3,0×10 K:-4.0 × 1010 Kal very large K201.2×10-2 K,-4.0 x 10 K,-1.0× 10 K-10 10 -60x10 HBO Kal-4.2 × 10" 10-1s x 102 " 5.9 × 102 K,-5.0×104 I. Calculate the pK, for a weak acid, HA, that is 2.3% ionized in 0.080 M solution ? a. 4.37 c. 1.66 d. 2.33 e. 3.09 2. Calculate the pH of 0.10 MHCN solution. K,-40x10 a. 6.75 b. 5.20 c. 8.42 d. 9.52 e. 10.4 L 3, wha 3. What is the percent ionization of 0.20 MHNO,? K,-45 x 10+ a. 1.0% b, 2.8% c. 4.6% d. 5.3% e. 5.9% 4. Assume that five weak acids, identified only by numbers (.,II,I,IV, and V), havethe ollowing ionization constants Acid Ionization Constant (K, value 1.0× 10-3 3.0 x 10 2.6 x 10-7 4.0 × 10-9 7.3 x 10-11 IV V. A 0.10 M solution of which acid would have the highest pH?

Explanation / Answer

As you know the answers

I am giving you the formula to calculate the following

1) We know that

Ka = concentration of acid X (degree of dissociaiton)^2

So Ka = 0.08 x (0.023 )^2

Ka = 4.23 X 10^-5

So pKa = -log[Ka] = 4.37

2) pH = -log[H+]

H+ = Concentration of Acid X degree of dissociation

Degree of dissociation = (Ka / Concentration of acid) ^1/2

so H+ = (Ka X concentration of acid ) ^2

H+ = (4 X10^-10 X 0.1 )^1/2 = 6.32 X 10^-6

so pH = 5.199

3) Percent ionization = = (Ka / Concentration of acid) ^1/2

% ionization = (4.5 X 10^-4 / 0.2 )^1/2 = 4.7 X 10^-1 = 4.6 %

4) highest pH will be of with lowest Ka value as lower the ka value lower the dissociation lower the H+ and higher the pH

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