What is the concentration of formic acid 0.10 M and its conjugate base at equilibrium? Ka = 1.7 x 10^-4. Please show work.?

HCOOH(aq) + H₂O(l) ⇌ HCOO⁻(aq) + H₃O⁺(aq) .... Kₐ

Initial concentrations :
[HCOOH]ₒ = 0.10 M
[HCOO⁻]ₒ = [H₃O⁺]ₒ = 0 M

Equilibrium concentrations :
Let [HCOO⁻] = [H₃O⁺] = y M
Then, [HCOOH] = (0.10 - y) M

Kₐ = [HCOO⁻] [H₃O⁺] / [HCOOH]
1.7 × 10⁻⁴ = y² / (0.10 - y)
(1.7 × 10⁻⁵) - (1.7 × 10⁻⁴)y = y²
y² + (1.7 × 10⁻⁴)y - (1.7 × 10⁻⁵) = 0
y = {-(1.7 × 10⁻⁴) ± √[(1.7 × 10⁻⁴)² + (1.7 × 10⁻⁵)]} / 2
Rejecting the negative answer, y = 4.0 × 10⁻³

[HCOOH] = {0.1 - (4.0 × 10⁻³)} M = 0.096 M
[HCOO⁻] = 4.0 × 10⁻³ M

HCOOH <--> H+ HCOO-

im not sure what you are asking here, the concentration of the H+ and HCOO- are .1?

if you have .1M of the H+ and HCOO- it will be X/(.1-x)^2 equal to 1.7x10^-4 and multiply that out and get a quadratic eequation. set it equal to 0 and do the quadratic formula. to make sure you get it right use one of the quadratic formula calculators on the web

if its the the other way around and you have .1M of formic acid and are going towards the H+ and HCOO- it will be

X^2/(.1-x) equal to 1.7X10^-4 multiply that out get the quadratic and solve for X. dont be scared to write 1.7x10^-4 as .00017 either. all you have to do is write it down and type it in the calculator

hope this helps