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

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:

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

$K_{w} = e^{-31.49} = 2.12 \cdot 10^{-14}$  

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!