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Lilit [14]
3 years ago
14

What is the molarity of a stock solution if 60 mL were used to make 150 mL of a .5M solution

Chemistry
1 answer:
kirza4 [7]3 years ago
6 0

Answer:

The answer to your question is 1.25 M

Explanation:

Data

Molarity 1 = ?

Volume 1 = 60 ml

Molarity 2 = 0.5 M

Volume 2 = 150 ml

Process

1.- Write the dilution formula

       Molarity 1 x Volume 1 = Molarity 2 x Volume 2

-Solve for Molarity 1

       Molarity 1 = Molarity 2 x Volume 2 / Volume 1

-Substitution

        Molarity 1 = (0.5)(150) / 60

-Simplification

        Molarity = 75 / 60

-Result

        Molarity = 1.25 M

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Explanation:

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What is the ph of a solution of 1.699 l of 1.25 m hcn, ka = 6.2 x 10-10, and 1.37 moles of nacn?
BlackZzzverrR [31]

The pH of a solution is 9.02.

c(HCN) = 1.25 M; concentration of the cyanide acid

n(NaCN) = 1.37 mol; amount of the salt

V = 1.699 l; volume of the solution

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pKa = - log (6.2 × 10⁻¹⁰)

pKa = 9.21

Henderson–Hasselbalch equation for the buffer solution:

pH = pKa + log(cs/ck)

pH = pKa + log(cs/ck)

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More about buffer: brainly.com/question/4177791

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