Determine the ka of an acid whose 0.294 m solution has a ph of 2.80. 1.2 x 10-5 8.5 x 10-6 2.7 4.9 x 10-7 5.4 x 10-3

1) Calculate [H+] from the pH:

pH = log { 1 / [H+] } = - log [H+]

=> [H+] = 10 ^ (-pH)

=> [H+] = 10 ^ (-2.80) = 0.00158

2) Assume the stoichiometry 1:1

=> HA aq ---> H(+) aq+ A(-) aq

=> [A-] = [H+] = 0.00158

[HA] = 0.294 – 0.00158 = 0.29242

3) Calculate Ka

Ka = [H+] *[A-] / [HA] = (0.00158)*(0.00158) / 0.29242 =8.54 * 10^ -6

Answer: 8.5 * 10^ -6

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okpalawalter8 okpalawalter8 · Answer 2 · 2022-02-21T14:26:01+01:00

We have that the acid dissociation constant(Ka) is mathematically given as

Ka=8.54e-6

The acid dissociation constant(Ka)

Question Parameters:

Generally the equation for the H+ pH value is mathematically

pH = log { 1 / [H+] }

[tex][H+] = 10 ^{-pH}[/tex]

[H+]= 0.00158

Where the stoichiometric ratio is 1:1

HA aq ---> H(+) aq+ A(-) aq

[H+] = 0.00158

[HA] = 0.294 – 0.00158

[HA]= 0.29242

Therefore

[tex]Ka =\frac{ [H+] *[A-]}{ [HA] }\\\\Ka=\frac{ (0.00158)*(0.00158)}{0.29242}[/tex]

Ka=8.54e-6

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