What is the ph of a solution of 0.20 m hno2 containing 0.10 m nano2 at 25°c

This is an incomplete question, here is a complete question.

What is the pH of a solution of 0.20 M HNO₂ containing 0.10 M NaNO₂ at 25°C, given Ka of HNO₂ is 4.5 × 10⁻⁴?

Answer : The pH of the solution is, 3.05

Explanation : Given,

for HNO₂ = 4.5 × 10⁻⁴

Concentration of = 0.20 M

Concentration of = 0.10 M

First we have to calculate the value of .

The expression used for the calculation of is,

Now put the value of in this expression, we get:

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

Now put all the given values in this expression, we get:

Therefore, the pH of the solution is, 3.05

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied