1. Determine the pH and pOH (to two decimal places) for each: (a) Pure water at

Question

1. Determine the pH and pOH (to two decimal places) for each: (a) Pure water at 50°C (K-5476 × 10-14) A: 6.63; 7.37 (b) An aqueous solution containing 3.50 × 10-3 M NaOH (at 250 A: 11.54; 246 (c) An aqueous solution containing 3.50 x 10-3 M H2SO4 (at 25) A: 215; 1185 2. Determine [H O*] and IOH ] concentrations (in M to two decimal places) for each: (a) Pure water at 10°C (Kw = 0.293 × 10-14) A: 541x10 M:541x10M (b) An aqueous solution with a pH of 3.23 A: 5.89 x10 M;1.70x10-11M Worksheet 14 Acid-Base Equilibria

Explanation / Answer

In order to be able to calculate pH, pOH and [H3O+]/[H+] and [OH-] we must understand the relationship between each of them

Remember that water will ionize as follows:

H2O(l) + H2O(l) <-> H3O+(aq) + OH-(aq)

so

Kw = [H3O+][OH-]

at T = 25°C, the Kw value is given as 10^-14

then,

10^-14 = [H3O+][OH-]

can be used always to relate H3O+ and OH-

also, note that

pH = -log([H+])

pOH = -log([OH-])

and

pH + pOH = 14 (stated above)

a)

Kw = [H+][Oh-]

Kw = x^2

x = sqrt(Kw)

[H+] = sqrt(5.476*10^-14) = 2.34*10^-7

pH = -log( 2.34*10^-7

ph = 6.630

b)

if

NaOH = [OH-] = 3.5*10^-3

pH= 14 - pOH = 14 + log(OH)

pH = 14 + log(3.5*10^-3)

ph = 11.54

c)

2[H2SO4] = [H+]

[H+] = 2*(3.5*10^-3) = 7.0*10^-3

pH = -log( 7.0*10^-3)

ph = 2.1549

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