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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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Explanation:

alright, dawg, lets get this bread. CHEMISTRY? OH YEAH I LOVE CHEMISTRY.

what is a mol? do you know who avogadro is? sounds like avocado. free shavocado. ok so you MUST REMEMBER THIS NUMBER PLEASE.

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now that we understand this..... lets perform this calculation with a calculator

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notice i divide the question by the avogadros number to find out how many moles are in the number. ok but listen... it gets into a tough area here... because HOW ARE WE TO DIVIDE SUCH A HUMONGOUS NUMBER BY ANOTHER HUMONGOUS NUMBER?!?!?

its easy, its cake, just listen this is how you do it. only focus on the numbers NOT the 10 exponential ones. so just 3.90 and 6.02 ok? lets divide these two numbers 3.90 / 6.02 and we get 0.6478... how interesting... ok now lets deal with the exponents of 10. notice that we are DIVIDING these numbers so think of it as subtracting the exponents of ten.....    22 minus 23 equals -1

so we have 0.6478 * 10^{-1}

now this negative 1 thing is annoying so lets just make it to the power of 0

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