Nitric acid (HNO3) is used in the production of fertilizer, dyes, drugs, and explosives. Calculate the pH of a HNO3 solution having a hydrogen ion concentration of 0.71 M.

pH of HNO₃ having an hydrogen ion concentration of 0.71M is 0.149

Explanation:

HNO₃ (aqueous) ⇄ H⁺ + NO3⁻

The pH is defined as the negative log of the hydrogen ion concentration

pH = - log [H⁺]

From the question, the hydrogen ion concentration is given as 0.71M, therefore

pH = -log [0.71]

     = 0.149

0.15

Explanation:

Whenever we are asked to calculate the pH of a particular solution in Chemistry it means that we are to look for its degree of acidity or basicity of such solution. Since the question is about an acidic solution, that is HNO3 we are to find the degree of acidity in the solution.

Therefore, the formula for the pH of a solution is given below;

pH = - log [ H^+].

In the question, we are given the hydrogen ion concentration to be = 0.71 M.

Hence, pH = - log [0.71].

pH = 0.14874165128092471.

pH =0.15.

It means that the solution is very acidic.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

"record the time to complete a chemical reaction" would most likely be part of a lab procedure.