if a solution measured with the miscalibrated meter (above) gave a reading of 6.50 at 20oC, what should be the true pH (give 2 d
1 answer:
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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⁺]
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
7665 years
Procedure
Let N₀ be the amount of carbon-14 present in a living organism. According to the radioactive decay law, the number of carbon-14 atoms, N, left in a dead tissue sample after a certain time, t, is given by the exponential equation:
N = N₀e^(-λt)
where λ is the decay constant which is related to half-life (T1/2) by the equation:
Here, ln(2) is the natural logarithm of 2.
The percent of carbon-14 remaining after time t is given by N/N₀.
Using the first equation, we can determine λt.
The half-life of carbon-14 is 5,720 years, thus, we can calculate λ using the second equation, and then find t.
Solving the second equation for t, and using the λ we have just calculated we will have
t= 7665 years