A solution of nitric acid is formed by dissolving 5.00grams of nitric acid in enough water to make 1.5 liters of solution. c. What was the molarity of this solution?

d. What is the pH of this solution?

Molarity = 0.053 M

pH = 1.276

Given

5 g of HNO₃

1.5 L solution

Required

Molarity

pH

Solution

Molarity shows the number of moles of solute in every 1 liter of solution.

mol HNO₃(MW=63 g/mol) :

= mass : MW

= 5 g : 63 g/mol

= 0.079

Molarity :

=0.079 : 1.5 L

= 0.053 M

pH = -log [H⁺]

For strong acids like HNO₃ :

[H⁺] = a.M

a = valence of acid / amount of H⁺ released  

M = acid concentration.  

HNO₃ ⇒ H⁺ + NO₃⁻⇒ valence 1(amount of H⁺=1)

[H⁺]=1 x 0.053

[H⁺]= 0.053 M

pH= - log 5.3 x 10⁻²

pH = 2 - log 5.3

pH= 1.276

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water at room temperature is flowing through a pipe with pressure taps located 10 ft apart along one section of the pipe, which are connected to a mercury (s=13.6) manometer.