Calculate the ph of a solution formed by mixing 200.0 ml of 0.30 m hocl (hypochlorous acid) with 300.0 ml of 0.20 m naocl (sodium hypochlorite). the ka for hocl is 2.9 × 10-8 .

The pH of the solution is 7.54

Explanation:

This is a buffer because we have a weak acid HOCl with its conjugate base OCl with Ka of 2.9×10⁻⁸ thus pKa = -log Ka = 7.54

therefore using Henderson Hasselbach equation: pH=pKa+log([OCl]/[HOCl]) we can determine the pH

n = mole

n of HOCl = C×V = 0.3 M ×0.2 L = 0.06 mole

n of NaOCl = 0.2 M × 0.3 L = 0.06 mole

Total volume = 300 + 200 = 500 mL = 0.5 L

final concentration: [HOCl] = 0.06/0.5 = 0.12 M; [OCl] = 0.06/0.5 = 0.12

pH=7.54+log(0.12/0.12) = 7.54

pH = pKa because the amount of HOCl = the amount of OCl in the final solution.

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

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explanation: