Question

1)What volume of a 15.0% by mass NaOH solution, which has a density of 1.116 g/mL , should be used to make 4.85 L of an NaOH solution with a pH of 10.5?

2)A 0.144 M solution of a monoprotic acid has a percent dissociation of 1.60%.Determine the acid ionization constant (Ka) for the acid.

3)A 9.5×10⁻² M solution of a monoprotic acid has a percent dissociation of 0.59%.Determine the acid ionization constant (Ka) for the acid.

Answer 1

ans)

1)

pH = 10.4

Thus, pOH = 14 - pH = 3.5

Thus, [OH^-] = 10^-3.5 = 3.162*10^-4M

Thus, [NaOH] = [OH^-] = 3.162*10^-4 M

moles of NaOH = molarity*volume of solution in litres = 0.0015

mass of NaOH = moles*molar mass = 0.0015*40 = 0.06 g

Thus, mass of 15% of NaOH solution = 0.06/0.15 = 0.4 g

Thus, volume of 15% NaOH solution required = mass/density = 0.4/1.116 = 0.358 ml

2)

from bove data that

   = 1.6%   = 1.6/100 = 0.016

= Ka/C

0.016 = Ka/0.144

(0.016)²   = Ka/0.144

Ka            = 0.016*0.016*0.144 = 3.69*10^-5 >>>> answer

3)

from bove data that

   = 0.59%   = 0.59/100 = 0.0059

= Ka/C

0.0059 = Ka/0.095

(0.0059)²   = Ka/0.095

Ka            = 0.0059*0.0059*0.095= 3.30*10^-6 >>>> answer