Question

(a) Calculate the quotient in $\left[H_3{\mathrm{PO}}_4\right]\mathbf{/}\left[H_3{{\mathrm{PO}}^{-}}_4\right]\mathbf{}\mathbf{in}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{0500}\mathbf{M}\mathbf{}{\mathbf{KH}}_{\mathbf{2}}{\mathbf{PH}}_{\mathbf{4}}$.

(b) Find the same quotient for $\mathbf{0}\mathbf{.}\mathbf{0500}{\mathbf{MK}}_{\mathbf{2}}{\mathbf{PO}}_{\mathbf{4}}$.

Short answer

(a) The quotient and pH of the proposed answer are 4.70 and $4.70\mathrm{and}2.80\times {10}^{-3}$,

respectively.

(b) The proposed solution's pH and quotient are 9.70 and $2.80\times {10}^{-8}$,

respectively.

Step-by-step solution

Definition of KH2PO4

• MKP also potassium dihydrogenphosphate, KDP, or monobasic potassium phosphate, KH2PO4, is a soluble potassium salt containing the dihydrogen phosphate ion.
• It's a good supplier of phosphate and potassium, as well as a good buffer.

Determine the supplied solution's pH and quotient

(a)

The quotient of a concentrated source ${\mathrm{KH}}_2{\mathrm{PO}}_4$It is necessary to compute the solution.

The pH of a solution can be determined using the formula:

localid="1655006932505" $\left[{\mathrm{H}}^{+}\right]=\frac{\sqrt{{\mathrm{K}}_1{\mathrm{K}}_2+{\mathrm{K}}_1{\mathrm{K}}_{\mathrm{w}}}}{{\mathrm{K}}_1+\mathrm{F}}$

The negative logarithm of Hydrogen ion activity can be calculated using the equation above

The pH and quotient of a given ${\mathrm{KH}}_2{\mathrm{PO}}_4$solution can be calculated as follows:

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]=\sqrt{\frac{{\mathrm{K}}_1{\mathrm{K}}_2(0.05000)+{\mathrm{K}}_1{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_1+(0.0500)}} \\ =1.99\times {10}^{-5} \\ \mathrm{pH}=4.70 \\ 4.70=2.148+\log \frac{\left[{\mathrm{H}}_2{\mathrm{PO}}_4\right]}{\left[{\mathrm{H}}_3{\mathrm{PO}}_4\right]}=\frac{\left[{\mathrm{H}}_3{\mathrm{PO}}_4\right]}{\left[{\mathrm{H}}_2{\mathrm{PO}}_4\right]} \\ =2.8\times {10}^{-3} \\ \mathrm{The}\mathrm{supplied}\mathrm{solution}\text{'}\mathrm{s}\mathrm{pH}\mathrm{and}\mathrm{quotient}\mathrm{are}\mathrm{both}4.70\mathrm{and}2.80\times {10}^{-3} \\ \mathrm{respectively}. \end{gathered}$$

Determine the proposed solution's pH and quotient

(b)

The quotient of a concentrated source ${\mathrm{K}}_2{\mathrm{HPO}}_4$It is necessary to compute the solution.

The pH and quotient of a given ${\mathrm{K}}_2{\mathrm{HPO}}_4$solution may be computed using the following formula:

$\left[{\mathrm{H}}^{+}\right]=\frac{\sqrt{{\mathrm{K}}_1{\mathrm{K}}_2+{\mathrm{K}}_1{\mathrm{K}}_{\mathrm{w}}}}{{\mathrm{K}}_1+\mathrm{F}}$

The negative logarithm of Hydrogen ion activity can be calculated using the equation above.

The pH and quotient of a given solution may be computed using the following formula:

role="math" localid="1655007416530" $$\begin{gathered} \left[{\mathrm{H}}^{+}\right]=\sqrt{\frac{{\mathrm{K}}_2{\mathrm{K}}_3(0.05000)+{\mathrm{K}}_2{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_2+(0.0500)}} \\ =1.99\times {10}^{-10} \\ p\mathrm{H}=9.70 \\ 9.70=22.148+\log \frac{\left[{\mathrm{H}}_2{\mathrm{PO}}_4\right]}{\left|{\mathrm{H}}_3{\mathrm{PO}}_4\right]} \\ =\frac{\left[{\mathrm{H}}_3{\mathrm{PO}}_4\right|}{\left[{\mathrm{H}}_2{\mathrm{PO}}_4\right]} \\ =2.80\times {10}^{-8} \end{gathered}$$

The proposed solution's pH and quotient are 9.70 and $2.80\times {10}^{-8}$ , respectively.