A 0.2 m aqueous solution of KCl freezes at -0.68°C. Calculate ‘i’ and the osmotic pressure at 0°C.
Answers
H3O + ] = 7.9 x 10-11 M
The pH can be calculated using the formula:
pH = -log10[H3O + ]
putting the value of hydronium ion concentration in above equation, we get:
pH = -log10[7.9 x 10-11]
pH = 11 - log10[7.9]
using the log table, we get:
pH = 11-0.8976 = 10.10
Since the pH of the solution comes out to be greater than 7, thus the solution is basic in nature.
2. [OH-] = 0.00424 M
The formula to calculate pOH is:
pOH = -log10[OH-]
Putting the value of hydroxide ion concentration in above equation:
pOH = -log10[0.00424]
= -log10[4.24 × 10-3]
= 3 - log10[4.24]
= 3 – 0.6274
= 2.37
Using the relation, pH + pOH = 14
pH = 14- 2.37 = 11.63
3. A) for solution of pH = 11.9, pOH = 2.1
We know that, pOH = -log10[OH-]
Putting given pOH value in above equation we get:
2.1 = -log10[OH-]
log10[OH-] = -2.1
log10[OH-] = 3 – 2.1 – 3
log10[OH-] = -3 + 0.9
[OH-] = antilog = 7.943 × 10-3 M
B) solution with pH = 8, pOH = 6
6 = - log10[OH-]
[OH-] = 10-6
Dividing the hydroxide ion concentration of s by Ist solution by second solution, we get: (7.943 × 10-3 ÷ 10-6 ) = 7943 times.
4. the formula used to find molarity is:
Putting the value of molarity = 0.2 M
Volume in mL = 500 mL
Molar mass = 98 g/mol
Weight = 9.8 g.
So, by adding 9.8 g of H2SO4 in the 500mL water, we can prepare its 0.2M solution.
5. the formula used to find molarity is:
Molarity = 0.03 M
Volume in mL = 125 mL
Molar mass of KOH = 56g/ mol
Weight = 0.21 g.
So by adding 0.21 g of KOH in 125 mL water, its 0.03 solution can be prepared.