Question

Chem 110B Name: Acid-Base Equilibria Pre-Lab 1. You have three beakers containing acidic solutions of the same volume. Beaker AS has a label of "0.1 M hydrochloric acid". Representative views of the acid molecules (water is omitted) in each beaker are below: C =HA O =A O=H a. Circle which of the following could be in beaker B. 0.1 M HNO 0.2 M HNO2 0.3 M HNO2 0.2 M HCl b. Explain your reasoning for part a. 0.3 M HCI C. is the solution in beaker C a strong or a weak acid solution? Briefly explain. 2. Ascorbic acid is also known as vitamin C. It has a K, of 8.0 x 109 at 25°C. Estimate the pH of a 0.100 M solution of ascorbic acid. What is the percent ionization of ascorbic acid?

Answer 1

Q.1.Ans :-

# Strong acids are the acids which are completely ionized as shown in becaker A in which black circles are not attached with white circles.

# Weak acids are the acids which are partially ionized as shown in becakers B and C in which black circles are attached with white circles.

(a). 0.2 M HNO2 could be in beaker B.

(b). Because beaker B contains a weak acid and HNO2 is a weak acid. 0.2 M HNO2 means the number of moles of HNO2 are 2 in 1L. Since, there are four black circles in beaker B which are denoted two black circles for HNO2 and two black circle forNO2-

(c). Beaker C contains a weak acid because black circles are attached with white circles.

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2. Ans :-

ICE table is :

............................HA (aq) <-----------------------------------> H+ (aq).........................+.........................A- (aq)

Initial....................0.100 M....................................................0 M......................................................0 M

Change...............-0.100 y ..................................................+ 0.100 y...............................................0.100 y

Equilibrium.........0.100(1-y) M.............................................0.100 y M................................................0.100 y M

Where, y = Amount dissociated per mole

Expression of Equilibrium constant i.e. Ka (which is equal to the product of the molar concentration of products divided by product of the molar concentration of reactants raise to power their stoichiometric coefficient when reaction is at equilibrium stage).

Ka = [H+].[A-] / [HA]

8.0 x 10-5 = (0.100 y)2 / 0.100 (1-y)

8.0 x 10-5 = 0.100 y2 / (1-y)

8.0 x 10-5 / 0.100 = y2 / (1-y)

8.0 x 10-4 (1-y) = y2

y2 + 8.0 x 10-4 y - 8.0 x 10-4 = 0

On solving

y = 0.0279

Therefore, percent ionization = 0.0279 x 100% = 2.79 %

[H+] = 0.100 x 0.0279 = 0.00279

pH = - log [H+]

pH = - log 0.00279

pH = 2.55

Therefore, pH = 2.55