Question

How many mL of concentrated hydrochloric acid should be added to 48.3 g of sodium nitrite to prepare 2.50 L of a buffer solution with a pH of 2.60?

Answer 1

Answer:

47 mL

Explanation:

The equilibrium of nitrous acid is:

HNO₂ ⇄ NO₂⁻ + H⁺ Ka = 4,5x10⁻⁴

If desire pH is 2,60 the [H⁺] concentration in equilibrium is:

[H⁺] = $10^{-2,6}$ =2,51x10⁻³ M

Initial molarity of sodium nitrite is:

43,8g × $\frac{1mol}{68,9953 g}$÷ 2,5 = 0,254 M

Thus, in equilibrium the concentration of chemicals is:

[NO₂⁻] = 0,254 - x

[HNO₂] = x

[H⁺] =  2,51x10⁻³ = Y-x Where Y is initial concentration.

Equilibrium formula is:

4,5x10⁻⁴ =  $\frac{[2,51x10^{-3}[0,254 - x] ]}{[x]}$

Solving, x = 0,215

Thus, initial [H⁺] concentration is:

0,215 + 2,51x10⁻³ = 0,2175 M

If total volume is 2,50 L:

2,50L ×$\frac{0,2175 mol}{L}$ = 0,5438 mol of HCl

As molarity of concentrated hydrochloric acid is 11,65 mol per liter:

0,5438 mol HCl ×$\frac{1L}{11,65}$ = 0,047 L ≡ 47 mL

I hope it helps!