Question

R-0.08206 (atm - L)(mol - K)= 8.314 J/(mol-K) pH = -log[H30) pOH = -log(OH) pX=-logX [HO'] = 10 Ph. K, EK.(RT) pH + pOH = 14 = pK+pKb [OH (H:0= 10-14 = K, * Ks (in water at 25°C) pH =pK, + log([base)/(acid]) pOH =pKy + log (acid]/[base]) Molar Solubility = (*+ m for a salt that dissociates into ions with coefficients of n and m - b b -4ac where 0 = ax + bx+c VA= A/ X=- 1) Which of the following solutions would have the second lowest pH? Assume all their concentrations are 0.10 M at 25 °C. The acid is followed by its K, value. A) HCIO, 2.9 x 10 B) HCN, 4,9 x 100 C) HO,C,Hs, 6.5 x 10 D) HF, 3,5 x 104 E) HCIO, 1.1 x 102 2) 2) Which of the following are true about systems at equilibrium? A) The concentrations of each component are, by definition, equal to each other. B) This symbol is used in a reaction where an equilibrium is present: C) Once a system reaches equilibrium the concentrations of each component do not change over time. D) Catalysts change the value of the equilibrium constant. E) The rates (speeds) of the forward and reverse reactions are not the same. 3) The reaction below has a Ke value of 3.64 10. What is the value of Kp for this reaction at 25 °C? HONO(g) = NO(g) + OH(g) A) 1.49 x 10 B) 8.90 x 102 C) 3.64 x 10- D ) 6.09x 10 E ) 2.17 4) Express the equilibrium constant K for the following reaction. CO2(g) + C(s,graphite) = 2 CO(g) [CO(g)] [CO(g)] B) (CO2(g) [CO,(s)[C(s)] [CO, (3)][C(s)] [CO(g) [CO(g)] [CO (8) [CO,(8)]

Answer 1

1. Option D

Since more the ka, more the acidity and lesser the pH.

Second highest ka is for HF ,hence second lowest pH

2. Option C, at equilibrium, concentrations don't change over time.

3. Option B

We know that, KP=KcRT^dn

dn= difference in the gaseous products and Reactants.

KP= 3.64x10*-3(0.0821x298)^2-1

=8.9x10^-2

4. Option D

Since equilibrium constant is the ratio of gaseous products and gaseous Reactant.

Solids won't contribute to equilibrium constant.