Question

From a plot of ln(Kw) versus 1/T (using the Kelvin scale), estimate Kw at 37°C, which is the normal physiological temperature. Kw = What is the pH of a neutral solution at 37°C?

Answer 1

A) Kw (37°C) = 2.12x10⁻¹⁴

B) pH (37°C) = 6.84

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:

`ln(K_(w)) = -6218.6 \cdot (1/(37 + 273)) - 11.426 = -31.49`

`K_(w) = e^(-31.49) = 2.12 \cdot 10^(-14)`

Therefore, Kw at 37°C is 2.12x10⁻¹⁴

B) The pH of a neutral solution is:

`pH = -log([H^(+)])` (2)

The hydrogen ion concentration can be calculated using the following equation:

`K_(w) = [H^(+)][OH^(-)]` (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):

`K_(w) = ([H^(+)])^(2)`

which gives:

`[H^(+)] = \sqrt {K_(w)}`

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`

Therefore, the pH of a neutral solution at 37°C is 6.84.

I hope it helps you!