You wish to make a buffer solution using an acid (HA) with a Ka of 8.90 x 10-6 and a salt of its conjugate base (Na...

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science chemistry

Answer 1

Answer #1

Sol.

As Conc. of HA = [HA] = 1.50 M

Conc. of NaA = [NaA] = 2.25 M

Ka = 8.90 × 10^-6

pKa = - log(Ka) = - log( 8.90 × 10^-6 ) = 5.05

So , Using Henderson - Hasselbalch equation ,

pH = pKa + log ( [NaA] / [HA] )

= 5.05 + log ( 2.25 / 1.50 )

= 5.23

Also , Hydronium Concentration at equilibrium

= [H3O+] = 10^-pH

= 10^-5.23

= 5.89 × 10^-6 M

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Answer 2

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