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

Respuesta :

From the calculations, the pH of the final solution is 9.04.

What is the pH of the buffer?

We can use the Henderson Hasselbach equation to obtain the final pH of the solution in terms of the pKb and the base concentration.

Number of moles of salt = 250/1000 L * 0.5 M = 0.125 moles

Number of moles of base = 150/1000 L * 0.5 M = 0.075 moles

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

Molarity of base = 0.075 moles/ 0.4 L = 0.1875 M

Molarity of salt = 0.125 moles/ 0.4 L = 0.3125 M

pOH = pKb + log[salt/base]

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

pOH = 4.74 + log[0.3125/0.1875 ]

pOH = 4.96

pH = 14- 4.96

pH = 9.04

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