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

47 mL

Step-by-step 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 ×
`(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 =
`([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 ×
`(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 ×
`(1L)/(11,65)` = 0,047 L ≡ 47 mL

I hope it helps!