Question

Phosphate buffers are important in regulating the$p\mathrm{H}$of intracellular fluids. If the concentration ratio of ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}$to $\text{HP}{{\text{O}}_{\text{4}}}^{\text{2 -}}$in a sample of intracellular fluid is $\text{1.1: 1}$what is $p\mathrm{H}$the of this sample of intracellular fluid?

${\text{H}}_{\text{2}}\text{P}{{\text{O}}_{\text{4}}}^{\text{-}}\text{(aq)}\rightleftharpoons \text{HP}{{\text{O}}_{\text{4}}}^{\text{2 -}}{\text{(aq) +H}}^{\text{+}}\text{(aq)}$

${\mathit{K}}_{\mathit{a}}{\text{=6.2×10}}^{\text{-8}}$

Short answer

The required$\mathrm{pH}$ of intracellular fluid is 7.17.

Step-by-step solution

Definition of pH

The pH is a measurement of how acidic or basic aqueous or other liquid solutions are. The phrase, which is frequently used in chemistry, biology, and agronomy, converts hydrogen ion concentrations, which typically vary from around 1 to 1014 gram-equivalents per litre, into numbers between 0 and 14.

Write the formula of pH :

A buffer is a salt solution of a mild acid (or base). The fundamental purpose of a buffer is to keep a solution's pH constant.

The Henderson-Hasselbalch equation, which is given below and underlined in red, connects the pH of an acidic buffer (HA) to the acid dissociation constant$K_a$.

$\text{HA}\rightleftharpoons {\text{H}}^{\text{+}}{\text{+A}}^{\text{-}}$

${\text{pH = pK}}_{\text{a}}\text{+ log}\frac{\left[{\text{A}}^{\text{-}}\right]}{\text{[HA]}}$

$\left[{\text{A}}^{\text{-}}\right]$is an acid dissociation constant, and ${\text{pK}}_{\text{a}}$is an acid dissociation constant. $\text{[H A]}$is a concentration of the acid, and $\text{[H A]}$is a concentration of the salt/conjugate base.

Calculate the value of pH

The ratio of${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}$${\text{HPO}}_{\text{4}}^{\text{2 -}}$is$\text{1:1}$.

${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\rightleftharpoons {\text{HPO}}_{\text{4}}^{\text{2 -}}{\text{+H}}^{\text{+}}$

Using the Henderson-Hasselbalch equation as a guide:

${\text{pH = pK}}_{\text{a}}\left({\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right)\text{+ log}\frac{\left[{\text{HPO}}_{\text{4}}^{\text{2 -}}\right]}{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]}$

$$\begin{gathered} K_a\left(H_2{\text{PO}}_4^{-}\right){\text{=6.2×10}}^{\text{-8}} \\ {\text{pK}}_a\text{=-log}\left(K_a\right) \\ \text{pH=-log}\left({\text{6.2×10}}^{\text{-8}}\right)\text{+log}\frac{1}{\text{1.1}} \end{gathered}$$

$$\begin{gathered} \text{pH = 7.207 - 0.0414} \\ \text{pH = 7.17} \end{gathered}$$

The provided sample of intracellular fluid containing a phosphate buffer has a $\text{pH is 7.17.}$

Therefore, the pH of intracellular fluid is 7.17.