Question

Problem Set 16 CHEM 1252 November 13, 2019 2. Consider the titration of 60.0 mL of 0.024 M NaOH with 0.036 M HCL write the net ionic equation for the titration reaction, including physical states. What p between this reaction and the k. (autoprotolysis) reaction? Given this relation the equilibrium constant for the titration reaction, and why can we safely assume sono completion"? Why should you expect the pH to equal 7.00 at the equivalence on (D) Calculate the volume of titrant required to reach the equivalence point. Include units throughout with an explicit mole ratio. (c) Calculate the initial pH. (d) Calculate initial mmol (millimoles) of OH-. (f) Calculate mmol H+ in 60.0 mL titrant. (e) Calculate mmol H. in 25.0 mL titrant. (9) Calculate pH after adding 25.0 mL titrant. (h) Calculate pH after adding 60.0 ml. titrant.

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Answer 1

2. a)

Molecular equation :

NaOH(aq) + HCl (aq)  $\to$ NaCl (aq) + H₂O(l)

Net ionic equation :

H⁺ (aq) + OH^- (aq)  $\to$ H₂O (l)

K_a = $\frac{1}{[H^{+}][OH^{-}]}$

Autoprotolysis reaction is

H₂O  $\to$ H⁺ + OH^- .

K_W = [H⁺][OH^-] = $\frac{1}{Ka}$

The neutralisation reaction goes to completion

as K_w = 10^-14 ,

hence, Ka = 10¹⁴ .

Now, at neutralisation point [H⁺] = [OH^-]

Or, [H⁺][OH^-] = 10^-14

Or, [H⁺]² = 10^-14

Or, [H⁺] = 10^-7 .

Or, pH = - log[H⁺] = - log10^-7 = 7.

b.

At equivalance point

Moles of NaOH = moles of HCl.

Or, M_NaOHV_NaOH = M_HClV_HCl

So, 60× 0.024 = x × 0.036

Or, x = (60×0.024/0.036) = 40 mL.

Hence, volume of titrant = 40 mL.

c.

Initial pH calculation.

pOH = - log [OH^-] = - log(0.024) = 1.62.

So, pH = (14 - 1.62) = 12.38

d.

Initial mmoles of OH^-

= Molarity of NaOH × volume of HCl

= 0.024×60

= 1.44

e)

mmoles of H⁺ in 25 mL

= Volume × molarity of HCl

= 25× 0.036

= 0.9.

f)

mmoles H⁺ in 60 mL titrant

= 60× 0.036

= 2.16

g)

mmoles of OH^-= 1.44

mmoles of H⁺ in 25 mL = 0.9

Excess mmoles of OH^- = 1.44 - 0.9 = 0.54.

Total volume = 60+25 = 85 mL

[OH^-] = (0.54/85) = 0.00635

Now,

pOH = - log [OH^-] = - log (0.00635) = 2.2

pH = 14 - 2.2 = 11.8.

h)

After adding 60 mL titrant (HCl)

mmoles of H⁺ = 2.16

mmoles of OH^- = 1.44

Excess of mmoles of H⁺= (2.16 - 1.44) = 0.72

Total volume = 60 + 60 = 120 mL

[H⁺] = (0.72/120) = 0.006

pH = - log[H⁺] = - log(0.006) = 2.22