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Montano1993 [528]
1 year ago
15

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)
Chemistry
1 answer:
Juli2301 [7.4K]1 year ago
5 0

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

<h3>What is the pH of the buffer?</h3>

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

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

#SPJ1

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