Strong base is dissolved in 625 mL of 0.400 M weak acid (Ka = 3.76\times10-5) to make a buffer...

Question:

Strong base is dissolved in 625 mL of 0.400 M weak acid {eq}(Ka = 3.76\times10-5) {/eq}to make a buffer with a pH of 4.09. Assume that the volume remains constant when the base is added.

Inorganic chemistry :

It is one of the part of chemistry which deals with the physical properties of the chemicals. In the inorganic chemistry the properties of the matter like volume, molarity etc are been studied and the theories behind this.

first, calculate {eq}pK_{a} {/eq} from {eq}K_{a} {/eq} value using the equation as:

{eq}pK_{a} = -log (K_{a}) = -log(3.76 \times 10^{-5}) = 4.4 {/eq}

{eq}pH = pK_{a} + log (\dfrac{base}{acid}) {/eq}

{eq}4.09 = 4.4 + log (\dfrac{base}{acid}) {/eq}

{eq}log (\dfrac{base}{acid}) = 4.09 - 4.4= -0.31 {/eq}

{eq}(\dfrac{base}{acid}) = 10^{-0.31} =0.500 {/eq}

{eq}moles \enspace of \enspace the \enspace acid = Molarity \times volume \enspace in \enspace liters = 0.4 \times 0.625L = 0.25 \enspace moles {/eq}

We can write chemical reaction as:

{eq}HA + XOH \rightarrow XA + H_{2}O {/eq}

so here no of moles of salt formed is equal to no of moles of a strong base

lets base = x

{eq}\dfrac{x}{0.25-x} = 0.5 {/eq}

{eq}x= 0.083M {/eq}

{eq}moles = M \times volume = 0.083 \times 0.625 L = 0.0518 \enspace Moles {/eq}