Verified answer

The equilibrium is

HBrO + H2O <=> H3O+ + BrO-

[H3O+]= 10^-4.48=3.31 x 10^-5 M = [BrO-]

Ka = [H3O+][BrO-]/ [HBrO] = ( 3.31 x 10^-5)(3.31 x 10^-5 )/ 0.55 =1.99 x 10^-9

The pH of a 0.55-M aqueous solution of hypobromous acid, HBrO, at 25°C is 4.48. What is the value of Ka for HBrO?

HBrO + H2O → H3O+1 + BrO-1

According to the equation above, 1 mole of HBrO molecules will produce 1 mole of H3O+1 ions and 1 mole of BrO-1 ions.

Ka = [H3O+1] * [BrO-1] ÷ [HBrO]

pH = - log [H3O+1]

[H3O+1] = 10^-4.48 = 3.311 * 10^-5

Since this reaction produced 3.311 * 10^-5 moles of H3O+1 ions per liter of solution, this reaction also produced 3.311 * 10^-5 moles of BrO-1 ions per liter of solution!

Leaving (0.55 – 3.311 * 10^-5) moles of HBrO molecules per liter of solution.

Ka = (3.311 * 10^-5)^2 ÷ (0.55 – 3.311 * 10^-5)

Ka =

you need to establish an ICE chart (I = preliminary, C = replace, E = equilibrium) for a susceptible acid. HB = cinnamic acid on the precise of the chart, you're able to have HB --> H+ + B- The preliminary quantity of HB is a million.6*10^-3 The replace for HB is -x, whilst the replace for H+ and B- is +x The equilibrium quantity for HB is a million.6*10^-3 - x The equilibrium quantity for H+ is x. Ka = [H+]^2/[HB] So Ka = x^2/(a million.6*10^-3 - x) because of fact x is the type of small selection, that's faraway from the denominator without inflicting plenty disturbance. so which you're left with Ka = x^2/a million.6*10^-3 After plugging in Ka and fixing for x, you get: x = 2.40 3*10^-4 = [H+] -log ([H+]) = pH pH = 3.61

Well, it depends..