Answer:
A) Kw (37°C) = 2.12x10⁻¹⁴
B) pH (37°C) = 6.84
Step-by-step explanation:
The following table shows the different values of Kw in the function of temperature:
T(°C) Kw
0 0.114 x 10⁻¹⁴
10 0.293 x 10⁻¹⁴
20 0.681 x 10⁻¹⁴
25 1.008 x 10⁻¹⁴
30 1.471 x 10⁻¹⁴
40 2.916 x 10⁻¹⁴
50 5.476 x 10⁻¹⁴
100 51.3 x 10⁻¹⁴
A) The plot of the values above gives a straight line with the following equation:
y = -6218.6x - 11.426 (1)
where y = ln(Kw) and x = 1/T
Hence, from equation (1) we can find Kw at 37°C:
[tex] ln(K_{w}) = -6218.6 \cdot (1/(37 + 273)) - 11.426 = -31.49 [/tex]
[tex] K_{w} = e^{-31.49} = 2.12 \cdot 10^{-14} [/tex]
Therefore, Kw at 37°C is 2.12x10⁻¹⁴
B) The pH of a neutral solution is:
[tex] pH = -log([H^{+}]) [/tex] (2)
The hydrogen ion concentration can be calculated using the following equation:
[tex] K_{w} = [H^{+}][OH^{-}] [/tex] (3)
Since in pure water, the hydrogen ion concentration must be equal to the hydroxide ion concentration, we can replace [OH⁻] by [H⁺] in equation (3):
[tex] K_{w} = ([H^{+}])^{2} [/tex]
which gives:
[tex] [H^{+}] = \sqrt {K_{w}} [/tex]
Having that Kw = 2.12x10⁻¹⁴ at 37 °C (310 K), the pH of a neutral solution at this temperature is:
[tex] pH = -log ([H^{+}]) = -log(\sqrt {K_{w}}) = -log(\sqrt {2.12 \cdot 10^{-14}}) = 6.84 [/tex]
Therefore, the pH of a neutral solution at 37°C is 6.84.
I hope it helps you!
