Your help would be appreciated on these.. I couldn\'t figurethem out on the clas
ID: 679174 • Letter: Y
Question
Your help would be appreciated on these.. I couldn't figurethem out on the class quiz. What mass of potassiumhypochlorite must be added to 450. mL of water to give a solutionwith pH = 10.20? [Ka(HClO) = 4.0 ×108] and also what is the pH of a solution preparedby mixing 100. mL of 0.0500 M HCl with 300. mL of 0.500 M HF?[Ka(HF) = 7.1 ×104]? Your help would be appreciated on these.. I couldn't figurethem out on the class quiz. What mass of potassiumhypochlorite must be added to 450. mL of water to give a solutionwith pH = 10.20? [Ka(HClO) = 4.0 ×108] and also what is the pH of a solution preparedby mixing 100. mL of 0.0500 M HCl with 300. mL of 0.500 M HF?[Ka(HF) = 7.1 ×104]?Explanation / Answer
ANSWER )
the reaction will takes place as follows :-
ClO- + H2O <----> HClO + OH-
So at equilibrium we can say that
[OH-] = [HClO]
As pH + pOH = 14
pOH = 14 - 10.20 =3.80
and pOH =- log (OH)
[OH-] = 10^-3.80 =1.584 *10-4 M
K = Kw/Ka = 10-14 / 4.0 x 10-8 =2.5 x 10-7
2.5 x 10-7 = [OH-][HClO] / [ClO-] = ( 1.584 x 10-4)2 / [ClO-]
[ClO-] = 0.1003 M
moles in 0.450 L = 0.1003 x 0.450 =0.04513 M
The Mass KClO is equal to = 0.045135 mol x 90.55 g/mol = 4.086 g
Answer 2) first we calculate the individual H+ concentration
[H+] due to HCL = 0.05*100/400 = 0.0125 M
[H+] due to HF = 0.0185*300/400 = 0.0139 M
Now we add the H+ concentration
[H+] = 0.0125+0.0139 = 0.0264 M
by applying the formula
pH = -log[H+] =
pH= -log (0.0264) = 1.578
Hence the pH comes out to be equal to =1.578
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