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dem82 [27]
3 years ago
8

Calculate the pka of hypochlorous acid. The ph of a 0.015 m solution of hypochlorous acid has a ph of 4.64.

Chemistry
1 answer:
o-na [289]3 years ago
5 0

Answer:

  • pKa = 7.46

Explanation:

<u>1) Data:</u>

a) Hypochlorous acid = HClO

b) [HClO} = 0.015

c) pH = 4.64

d) pKa = ?

<u>2) Strategy:</u>

With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.

<u>3) Solution:</u>

a) pH

  • pH = - log [H₃O⁺]

  • 4.64 = - log [H₃O⁺]

  • [H_3O^+]= 10^{-4.64} = 2.29.10^{-5}

b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)

c) Equilibrium constant: Ka =  [ClO⁻] [H₃O⁺] / [HClO]

d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M

e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M

f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46

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A 25.0 ml sample of 0.150 m butanoic acid is titrated with a 0.150 m naoh solution. what is the ph after 26.0 ml of base is adde
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First,
Moles of HBu = Volume per liter * moles of HBu/ liter
                        =  25ml/1000 * 0.15 = 0.00375 moles
moles of NaOH = volume per liter * moles of NaOH/liter
                           = 1 ml/1000 * 0.15 =  0.00015 moles
according to this equation:
HBu + NaOH → H2O + NaBu
when 1 mol of NaOH gives 1mol of HBu
So     0.00015 of NaOH will give 0.00015 mol of HBu 
∴moles of HBu which remains =   0.00375- 0.00015 = 0.0036 moles 

∴moles of Bu- produced = 0.00015 moles
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∴[H+] =1x10^-6 / 0.076 = 1.5 x 10^-5
∴PH = -㏒[H+]
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        = 4.81

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