Solve for [H3O+] C2H3O2 = .0951 Hc2H3O2 = 0.1 H3O = x I know the formula is x (0

Question

Solve for [H3O+] C2H3O2 = .0951 Hc2H3O2 = 0.1 H3O = x I know the formula is x (0.0951 + x) / (0.1 - x) = 1.8 times 10-5 Im just not sure how to get x

Explanation / Answer

x(0.0951 + x) / (0.1 - x) = 1.8 x 10^-5 --> Now, just cross multiply x(0.0951 + x) = (1.8 x 10^-5)(0.1 - x) 0.0951x + x^2 = (1.8 x 10^-6) - (1.8 x 10^-5 x) x^2 + 0.0951x + (1.8 x 10^-5 x) - (1.8 x 10^-6) = 0 x^2 + 0.0951x - (1.8 x 10^-6) = 0 --> quadratic equation in the form ax^2 + bx + c = 0 a = 1 b = 0.0951 c = -1.8 x 10^-6 Use the quadratic formula [-b (+ or -) square root of b^2 - 4ac] / 2a to solve for x: You should get two numbers. Disregard the negative one. Hope this helps! :-)

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