Question

The buffer is prepared by adding 250mL of 0.80M NH3 to 150mL of 0.50M NH4NO3. What is the pH of the final solution (Kb for NH3 = 1.8 x 10^-5)

Answer 1

The pH of the solution is obtained to be 9.67.

What is a buffer?

A buffer is a solution that serves to militate against changes in acidity or alkalinity. The buffer is often made of a solution of a strong acid and its salt or a weak base and its salt.

Now we have the that;

Number of moles of base = 250 /1000 *  0.80 = 0.2 moles

Number of moles of salt = 150/1000 *  0.50 = 0.075 moles

Total volume of the solution = 250 + 150 = 400 mL or 0.4 L

Molarity of the base = 0.2 moles/0.4 L = 0.5 M

Molarity of the salt =  0.075 moles/ 0.4 L = 0.1875 M

pKb = - log( 1.8 x 10^-5) = 4.74

pOH = pKb + log [salt/base]

pOH =  4.74 + log (0.1875/ 0.5)

pOH = 4.33

pH = 14 - 4.33

pH = 9.67

The pH of the solution is obtained to be 9.67.

Learn more about pH:https://brainly.com/question/15289741

#SPJ1