Question

A solution is prepared by dissolving 106.3 g HCl(g) in enough water to make 175.0 L of solution. The ph of this solution is

Answer 1

Answer:

moles HCl = 101.2 g /36.461 g/mol=2.78

[H+]= 2.78 mol/ 150 L =0.0185

pH = - log 0.0185=1.73

1.73 is the answer

Answer 2

The pH of the solution resulting from the dissolution of 106.3 g HCl in 175.0 L of water is 1.77.  

The pH of a solution is given by:

$pH = -log([H_{3}O^{+}])$  (1)

Since HCl is a strong acid, we have that:

$[H_{3}O^{+}] = [HCl]$

We can calculate the concentration of HCl from its mass and the volume of solution:

$[HCl] = \frac{106.3 g}{175.0 L}*\frac{1 mol}{36.46 g} = 0.017 g/mol$

Hence, the pH is (eq 1):

$pH = -log([H_{3}O^{+}]) = -log(0.017) = 1.77$      

Therefore, the pH of the solution is 1.77.

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