Question

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?

Answer 1

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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