Question

Find ${\mathbf{pAg}}^{\mathbf{+}}$49.1 mL.

Short answer

The${\mathrm{pAg}}^{+}$value for the titration of${\mathrm{AgNO}}_3\mathrm{VsKI}$ is 12.86.

Step-by-step solution

Definition of molarity

• The ratio of moles of solute (in grammes) to the volume of the solution is known as molarity (in litres).
• The amount of a material in a given volume of solution is measured in molarity (M).
• The mole of a solute per litre of a solution is known as molarity.
• The molar concentration of a solution is also known as molarity.

Determine the pAg+ value for the titration of AgNO3VsKI

The ${\mathrm{pAg}}^{+}$titration value for the titration of${\mathrm{AgNO}}_3\mathrm{VsKI}$ It must be calculated.

$\mathrm{pX}=-{\log }_{10}\left[\mathrm{X}\right]$

The ratio of moles of solute (in grammes) to the volume of the solution is known as molarity (in litres). The formula for calculating a solution's molarity is:

$\mathrm{Molarity}\left(\mathrm{M}\right)=\frac{\mathrm{Molesofsolute}\left(\mathrm{ing}\right)}{\mathrm{Volumeofsolution}\left(\mathrm{in}\mathrm{L}\right)}$

The${\mathrm{pAg}}^{+}$ for the titration of value before equivalence point${\mathrm{AgNO}}_3\mathrm{VsKI}$

Volume of silverion at equivalency point is given=50.00mL

The reaction of titration of ${\mathrm{AgNO}}_3\mathrm{VsKI}$is the following:

${\mathrm{Ag}}^{+}+{\mathrm{I}}^{-}\to \mathrm{AgI}\left(\mathrm{s}\right)$

Silver ion and iodide ion solution total volume is$0.0741\mathrm{L}(25.00\mathrm{mL}+49.10\mathrm{mL})$

The iodide ion concentration is

$$\begin{gathered} \left[{\mathrm{I}}^{-}\right]=\left(\frac{0.9}{50.00}\right)\left(0.1000\mathrm{M}\right)\left(\frac{25.00\mathrm{mL}}{74.10\mathrm{mL}}\right) \\ =6.072\times {10}^{-4}\mathrm{M} \end{gathered}$$

Determine the silver ion concentration

The silver ion concentration that is in equilibrium with the iodide ion is

$$\begin{gathered} \left[{\mathrm{Ag}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{qp}}}{||\mathrm{T}||} \\ =\frac{8.3\times {10}^{-17}}{6.072\times {10}^{-4}\mathrm{M}} \\ =1.367\times {10}^{-13}\mathrm{Mp} \\ {\mathrm{Ag}}^{+}=-{\log }_{10}\left[{\mathrm{Ag}}^{+}\right] \\ =-\log \left(1.367\times {10}^{-13}\right) \\ =12.86 \end{gathered}$$

The ${\mathrm{pAg}}^{+}$value for the titration of ${\mathrm{AgNO}}_3\mathrm{VsKI}$was calculated.