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coldgirl [10]
1 year ago
9

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?

Chemistry
1 answer:
BlackZzzverrR [31]1 year ago
8 0

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

c(NaCN) = 1.37 mol ÷ 1.699 l

c(NaCN) = 0.806 M; concentration of the salt

Ka = 6.2 × 10⁻¹⁰; acid constant

pKa = -logKa

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)

pH = 9.21 + log (0.806M/1.25M)

pH = 9.21 - 0.19

pH = 9.02; potential of hydrogen

More about buffer: brainly.com/question/4177791

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3 years ago
Molar mass c3h8
maksim [4K]

Answer:

1. 44.11 g

2. 36.03 g

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4. 81.7%

5. 18.3%

Explanation:

1. 12.01+12.01+12.01+1.01+1.01+1.01+1.01+1.01+1.01+1.01+1.01=44.11

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<u>Explanation:</u>

We are given:

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Mass of NaCl, m_2 = (M-m_1)=(39.664-21.577)g=18.087g

To calculate density of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

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