Answer :

1. In distilled water: [H3O + ] × [OH-] = 10-14

For solution A, [H3O + ] = 10-14 ÷ [OH-]


= 10-14 ÷ (4.3 × 10-4)


= 2.32 × 10-11 M


For solution B, [H3O + ] = 7.3 × 10-10 M


Since the hydronium ion concentration of solution B is more, so its pH will be less.


The solution having the more pH will be more basic. As solution A will have more pH so it will be more basic.


2. using the relation, pOH + pH = 14


pOH = 14 – pH


= 14 – 9.3


= 4.7


Using, pOH = -log10[OH-]


4.7 = -log10[OH-]


log10[OH-] = -4.7


log10[OH-] = 5 -4.7 -5


log10[OH-] = -5 + 0.3


log10[OH-] = + 0.3


[OH-] = antilog


[OH-] = 1.99 × 10-5


3. molar mass of NaOH = atomic mass of Na + Atomic mass of O + atomic mass of H = 23 + 16 + 1 = 40g/ mol.


Volume of solution = 5 litres


Given weight = 8g


No. of moles = given weight ÷ molar mass


= 8g ÷ 40 g/mol


= 0.2 mole


Molarity = no. of moles ÷ volume in litres


= 0.2 ÷ 5


= 0.04 M


Therefore, the hydroxide ion concentration [OH-] = 0.04 M


pOH = -log10[OH-]


pOH = -log10[ 4 × 10-2]


pOH = 2 -log10[ 4]


pOH = 2- 0.6020 = 1.39


using, pOH + pH = 14


pH = 14 – pOH


pH = 14 - 1.39 = 12.61


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