A75.0-ml volume of 0.200 m nh3 (kb=1.8×10−5) is titrated with 0.500 m hno3. calculate the ph after the addition of 19.0 ml of hno3. express your answer numerically.

pH after the addition of 19.0 ml of HNO₃ : 9.018

The pH value of a reaction between strong acid HNO₃ and weak base NH₃ can be estimated from the rest of the reaction product

1. If the remainder of the reaction results obtained the remaining strong acid HNO₃, the pH is sought from the concentration of [H⁺] using the formula

[H⁺] = a. M

a = valence of acid / amount of H⁺ released

M = acid concentration

2. When strong acids and weak bases react, they form salts that are acidic, calculating the pH using the pH hydrolysis formula

where

M = concentration of salt anion

3. If the remainder of the reaction results are a weak base remaining and the salt, the solution will form a base buffer solution and search for pH using the formula base buffer pH

We count the moles of each reactant:

NH₃ mole = 75 ml x 0.2 M = 15 mlmol

mole HNO₃ = 19 ml x 0.5 M = 9.5 mlmol

NH₃ + HNO₃ ---> NH₄NO₃

15        9.5    

9.5      9.5            9.5

5.5       0              9.5

so there are remaining weak bases NH₃ = 15 - 9.5 = 5.5 mmol

Then a buffer solution is formed

[OH⁻] = Kb x [weak base mole] / [salt mole x valence]

the pH of a solution

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Calculate the pH

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the pH of a 2.0 M solution of HClO4

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Keywords : pH, acid, base, HNO₃,NH₃

the equation that correctly represent the reaction for formation of ammonia is

n2+ 3h2 → 2nh3 (answer d)

1 mole of nitrogen gas (n2) react with 3 moles of hydrogen gas (h2) to form 2 moles of ammonia ( nh3). this is in a process known as haber process were iron is used as a catalyst and reaction take place that a higher temperature and pressure. the process is exothermic hence energy is released.

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