Answer:
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
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
Answer: 1.14 M
Explanation:
500 mL is equal to 0.5 L.
So, the molarity is (0.570)/(0.5) = 1.14 M
The amount of HCl required for one experiment - 13.5 µl
the volume in terms of L - 13.5 x 10⁻⁶ L
the volume of HCl available - 0.250 L
since one experiment uses up - 13.5 x 10⁻⁶ L
then number of experiments - 0.250 L / 13.5 x 10⁻⁶ L = 1.8 x 10⁴ times
the experiment can be carried out 18000 times