Question

2L of HCl of pH 3 mixed with 500ml of Ba(OH)2 of pOH 3. Find the pH of mixture.
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Answer 1

Answer:

What would be the pH of solutions obtained by mixing 100 ml of 0.02 m HCl and 200 ml of 0.01 m Ba(OH)3?

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2HCl + Ba(OH)2 ————-> BaCl2 + 2H2O

Number of moles of HCl added= 0.02 x 10–3 molcm-3 x 100cm3 = 2 x 10–3 mol

Number of moles of Ba(OH)2 added= 0.01 x 10–3 molcm-3 x 200cm3 = 2 x 10–3 mol

Considering limiting reagents,

Number of moles of Ba(OH)2 needed for complete neutralization of HCl = 1 x 10–3 mol of Ba(OH)2

Number of moles of HCl needed for complete neutralization of Ba(OH)2 = 4 x 10–3 mol of HCl

Thus HCl is the limiting factor. So all the HCl get used up after the neutralization reaction. Ba(OH)2 however is in excess so the medium has to be basic.

Ba(OH)2 ————-> Ba2+ + 2OH-

Number of moles of excess Ba(OH)2= (2 x 10–3 mol) - (1 x 10–3 mol) = 1 x 10–3 mol

Number of moles of OH-= 2 x 1 x 10–3 mol = 2 x 10–3 mol

Concentration of OH-= ( 2 x 10–3 mol/ 300 cm3) x 1000 = 0.67 x 10–2 moldm-3

pOH = -log[OH-]

pOH = -log[0.67 x 10–2 moldm-3] = 2.1739

Considering the experiment was done at room temperature where Kw = 1 x 10–14 mol2dm-6

pH + pOH = 14

pH + 2.1739 = 14

pH= 11.8261 ( This agrees with our earlier confirmation that the medium should be basic)

Hope this will help you. ☺