Question

25 mL of 0.080 M solution of ammonia was titrated with 0.10 M HCl solution until the pH attained "saturation value". Draw the titration curve illustrating dependence between pH and volume of the HCI solution. Give the required answers in the boxes provided. Ko of NH3 = 1.8 x 10-5 Give initial pH value (must be correct within 0.5 pH unit) .... pH Circle whether the pH value at the equivalence point is lower than 7, equal to 7 or higher than 7. Give final pH value (correct within 0.5 pH unit)... Give correct shape and mark all inflection points (one or two) in appropriate regions of pH. Give the volume of HCl solution in ml (correct within 1 mL) required to reach the equivalence point VHCI [ML]

Answer 1

.30 на T

Initial pH :

[OH^-] = √ c.K_b = √ (0.080 × 1.8×10^-5) = 0.0012 M

pOH = - log (0.0012) = 2.921

pH = 14 - 2.921 = 11.079

pH = 11.08

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pH at equivalent point would be lower than 7 .

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Concentration of NH₄Cl = 0.080M×25ml/45ml = 0.0444M

pH = 7 - 1/2 × (pKb + log c) = 7 - 1/2 × (-log(1.8×10^-5) + log0.0444)

= 7 - 1.696 = 5.304

pH = 5.30 (pH at equivalent point)

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M1V1 = M2V2

0.080M × 25ml = V₂ × 0.10M

V₂ = 20ml

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