Question

What mass of hi should be present in 0.250 l of solution to obtain a solution with each ph value 2.85?

Answer 1

pH = 1.30  

[H+]= 10^-1.30 = 0.050  

HI is a strong acid so [HI] = 0.050 M  

Moles HI = 0.050 x 0.250 =0.0125  

Moles HI = 127.9 g/mol x 0.0125 mol = 1.60 g

Answer 2

Answer:

The mass of hi should be present in 0.250 L of solution is 0.04520 g.

Explanation:

The pH of the solution= 2.85

$pH=-\log[H^+]$

$2.85=-\log[H^+]$

$[H^+]=0.0014125 M$

1 mole of HI gives 1 mole of hydrogen ion and 1 mole of iodide ions.

$[H^+]=[HI]=0.0014125 M$

$Molarity =\frac{Moles}{Volume (L)}$

$0.0014125 M=\frac{n}{0.250 L}$

n = 0.0014125 M × 0.250 L = 0.00035 mol

Mass of 0.00035 moles of HI:

0.00035 mol × 128 g/mol = 0.04520 g