Answer :

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


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