Determine the pH of each of the following solutions (K_a and K_b values can be f

Question

Determine the pH of each of the following solutions (K_a and K_b values can be found in the appendices of you textbook) a. 0.045 M hypochlorous acid b. 0.0068 M phenol c. 0.080 M hydroxylamine Ephedrine, a central nervous stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base: C_10H_15NO(aq)+ H_2O(l) C_10H_15NOH^+(aq) OH^- (aq) K_b = 1.4 times 10^-4 What pH would you expect for a 0.035 M solution of ephedrine, assuming that no other substances are present? What is the value of pK_a for the conjugate acid of ephedrine? A buffer is prepared by adding 5.0 g of ammonia, NH_3, and 20.0 g of ammonium chloride, NH_4Cl, to enough water to form 2.50 L of solution. a. What is the pH of the buffer? b. a complete ionic equation for the reaction that occurs when a few drops of nitric acid are added to the buffer. c. Write the complete ionic equation for the reaction that occurs when a few drops of potassium hydroxide solution are add to the buffer.

Explanation / Answer

(a) 0.045 M hypochlorous acid

The formula of hypochlorous acid is HClO; we write the equilibrium reaction:

HClO(aq) H+(aq) + ClO-(aq)

and the expression for Ka:

Ka=[H+][ClO-]/[HClO] = 3*10^-8

As we know here is the initial concentration of HClO; we set up an ICE table, and we let x represent the amount of HClO which ionizes. this is also the amount of H+ and ClO- formed:

HClO(aq) =H+(aq) + ClO-(aq)

Initial 0.045 0 0

Change -x x x

Equilibrium 0.045-x x x

Let us plug the equilibrium concentrations into the expression for Ka; we obtain

X^2/ (0.045 - x) = 3.0*10^-8

We will first make the approximation that x is small relative to 0.045, so that 0.045 - x 0.045,

X^2/0.045 = 3.0 * 10^-8

We solve for x: x = Sq((3.0*10^-8)*(0.045)) = 3.67*10^-5

pH = -log(x) = -log (3.67*10^-5) = 4.43

(b)0.0068M Phenol

Phenol is HC6H5O

Its dissociation reaction is HC6H5O(aq) H+(aq) + C6H5O-(aq)

Initial 0.0068 0 0

Change -x x x

Equilibrium 0.0068-x x x

Ka of Phenol is = 1.3*10^-10

Using the approach same as above.

We solve for x: x = Sq((1.3*10^-10)*(0.0068)) = 9.40*10^-7

pH = -log(x) = -log (9.40*10^-7) = 6.02

C) 0.080 M hydroxylamine

As we know hydroxylamine is weak base, its reaction will be

HONH2(aq) + H2O(l) HONH3+(aq) + OH-(aq)

And Kb = 1.1*10^-8

Kb = [HONH3+][OH-]/[HONH2]

Initial 0.080 0 0

Change -x x x

Equilibrium 0.080-x x x

We will not consider water in calculations.

X^2 /(0.08 – x)= 1.1*10^-8

Using the approach same as above.

We solve for x: x = Sq((1.1*10^-8)*(0.080)) = 2.97*10^-5

As we have value of Kb we will get pOH

pOH = -log(x) = -log (2.97*10^-5) = 4.52

pH = 14-pOH= 14- 4.52 = 9.48

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