Solve the f

1. pH = 11.65

Therefore, pOH = 14 – 11.65 = 2.35

Using, pOH = -log₁₀[OH^-]

Putting the value of pOH in above equation:

2.35 = -log₁₀[OH^-]

log₁₀[OH^-] = -2.355

log₁₀[OH^-] = 3 -2.35 -3

log₁₀[OH^-] = -3 + 0.65

log₁₀[OH^-] = + 0.65

[OH^-] = antilog

[OH^-] = 4.46 × 10^-3

Now if volume is made 6 times then the concentration will decrease 6 times. So the new concentration is = 4.46× 10^-3÷ 6

= 7.44 × 10^-4

So new pOH = -log₁₀[OH^-]

= -log₁₀[7.44 × 10^-4]

= 4-log₁₀[7.44]

= 4- 0.871

= 3.129

Therefore, new pH is 14 – 3.129 = 10.871

2. A) when concentration is 0.03 M

Using, pH = -log₁₀[H₃O ⁺ ]

= -log₁₀[0.03]

= -log₁₀[3 × 10^-2]

= 2 -log₁₀[3]

= 2 – 0.477

= 1.52

B) When concentration is 0.05 M

Using, pH = -log₁₀[H₃O ⁺ ]

= -log₁₀[0.05]

= -log₁₀[5 × 10^-2]

= 2 -log₁₀[5]

= 2 – 0.699

= 1.30

So the change in pH is = 1.52 – 1.30

= 0.22