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?

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⁺] = =2,51x10⁻³ M

Initial molarity of sodium nitrite is:

43,8g × ÷ 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 =  

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 × = 0,5438 mol of HCl

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

0,5438 mol HCl × = 0,047 L ≡ 47 mL

I hope it helps!

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