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)
<em>where y = ln(Kw) and x = 1/T </em>
Hence, from equation (1) we can find Kw at 37°C:
![ln(K_{w}) = -6218.6 \cdot (1/(37 + 273)) - 11.426 = -31.49](https://tex.z-dn.net/?f=%20ln%28K_%7Bw%7D%29%20%3D%20-6218.6%20%5Ccdot%20%281%2F%2837%20%2B%20273%29%29%20-%2011.426%20%3D%20-31.49%20)
Therefore, Kw at 37°C is 2.12x10⁻¹⁴
B) The pH of a neutral solution is:
(2)
The hydrogen ion concentration can be calculated using the following equation:
(3)
<u>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):</u>
![K_{w} = ([H^{+}])^{2}](https://tex.z-dn.net/?f=%20K_%7Bw%7D%20%3D%20%28%5BH%5E%7B%2B%7D%5D%29%5E%7B2%7D%20)
which gives:
![[H^{+}] = \sqrt {K_{w}}](https://tex.z-dn.net/?f=%20%5BH%5E%7B%2B%7D%5D%20%3D%20%5Csqrt%20%7BK_%7Bw%7D%7D%20)
Having that Kw = 2.12x10⁻¹⁴ at 37 °C (310 K), the pH of a neutral solution at this temperature is:
![pH = -log ([H^{+}]) = -log(\sqrt {K_{w}}) = -log(\sqrt {2.12 \cdot 10^{-14}}) = 6.84](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%20%28%5BH%5E%7B%2B%7D%5D%29%20%3D%20-log%28%5Csqrt%20%7BK_%7Bw%7D%7D%29%20%3D%20-log%28%5Csqrt%20%7B2.12%20%5Ccdot%2010%5E%7B-14%7D%7D%29%20%3D%206.84%20)
Therefore, the pH of a neutral solution at 37°C is 6.84.
I hope it helps you!