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

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6