Please answer all. A 0, 109 M solution (25.00 ml) of ammonia (pK_0 = 4.76) is ti
ID: 526040 • Letter: P
Question
Please answer all.
A 0, 109 M solution (25.00 ml) of ammonia (pK_0 = 4.76) is titrated with 0, 100 M hydrochloric acid, Calculate the pH of the solution after the addition of 8.18 mL acid solution. What mass of sodium acetate must be added to 1.00 L of 0, 10 M acetic acid (pk_a= 4.75) to give a solution with a pH of 4. 54? Atomic masses: C 12.01, H 1, 01, 016.00, Na 22.99 A 5.98 %w/v solution (5.98g per 100.0 mL) of starch in water was found to have an osmotic pressure of 85.7 mmHg at 281.5 kelvin, Calculate the average molar mass of the starch used, taking 1,00 atm = 760.0 mmHg. The value of the gas constant, R_i is 0.0821 L atm K^-1 mol^-1.Explanation / Answer
1) millimoles of NH3 = 25 x 0.109 = 2.725
millimoles of HCl added = 8.18 x 0.1 = 0.818
2.725 - 0.818 = 1.907 millimoles NH3 left
0.818 millimoles salt formed
tota volume = 25 + 8.18 = 33.18 mL
[NH3] = 1.907 / 33.18 = 0.057 M
[NH4Cl] = 0.818 / 33.18 = 0.025 M
pOH = pKb + log [NH4Cl] / [NH3]
pOH = 4.76 + log [0.025] / [0.057]
pOH = 4.40
pH = 14 - 4.40
pH = 9.60
2) pH = pKa + log [sodium acetate] / [acetic acid]
4.54 = 4.75 + log [sodium acetate] / [0.1]
-0.21 = log [sodium acetate] + 1
log [sodium acetate] = -1.21
[sodium acetate] = 0.062 M
Molarity = (W/MW) (1 /V in L)
0.062 = (W / 60.05) (1/1)
W = 3.72 g