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

8.5 * 10^ -6

10 years is the answer

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.