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31 December, 06:44

Calculate the ph of a 0.170 m nh3-0.187 m nh4cl buffer when

a. 20.0 ml of 0.100 m hcl,

b. 20.0 ml of 0.100 m naoh, are added to 200 ml of buffer. ka nh4 + = 5.69 x 10-10

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  1. 31 December, 09:16
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    (a) 9.11; (b) 9.26

    pH of original buffer

    Step 1. Write the chemical equation for the equilibrium

    Acid + H_2O ⇌ base + H_3O^ (+); K_a = 5.69 * 10^ (-10)

    Step 2. Calculate the pH of the buffer

    The Henderson-Hasselbalch equation is pH = pK_a + log ([base]/[acid]).

    pH = - log[5.69 * 10^ (-10) ] + log[ (0.170 mol) / L) / (0.187 mol/L) ] = 9.24 + log0.1091

    = 9.24 - 0.0414 = 9.20

    (a) pH after adding HCl

    Step 1. Calculate the moles of acid and base in the buffer

    Moles of acid = 200 mL * (0.187 mmol/1 mL) = 37.4 mmol

    Moles of base = 200 mL * (0.170 mmol/1 mL) = 34.0 mmol

    Step 2. Calculate the moles of HCl added

    Moles of HCl added = 20.0 mL * (0.100 mmol/1 mL) = 2.00 mmol

    Step 3. Calculate the new moles of acid and base

    New moles of acid = (37.4 + 2.00) mmol = 39.4 mmol

    New moles of base = (34.0 - 2.00) mmol = 32.0 mmol

    Step 4. Calculate the new pH

    pH = 9.20 + log[ (32.0 mmol) / (39.4 mmol) ]) = 9.20 + log0.8122 = 9.20 - 0.0903 = 9.11

    (b) pH after adding NaOH

    Step 1. Calculate the moles of NaOH added

    Moles of NaOH added = 20.0 mL * (0.100 mmol/1 mL) = 2.00 mmol

    Step 2. Calculate the new moles of acid and base

    Moles of acid = (0.187 - 0.0200) = 0.167 mol

    Moles of base = (0.170 + 0.0200) = 0.190 mol

    Step 3. Calculate the new pH

    pH = 9.20 + log[ (0.190 mol/L) / (0.167 mol/L) ]) = 9.20 + log1.138 = 9.20 + 0.0560 = 9.26
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