# Ionic equilibrium

• a c d for 7th question mg concentration is .08. so OH is 4.47*10^-6. so ans is c.
8 is correct d as @Harshal-Patil has done it in continuation to 7th but it is in continuation to orginal question.

mg(cl)2 + 2naoh = 2nacl + mg(oh)2

so 0.1 mole of mg(cl)2 and remains and 0.1 mole of mg(oh)2 is formed

naoh is consumed

now we have basic buffer

so mgcl2 will give 0.1 mole ions of mg2+ and mg(oh)2 will give 2x ions of oh- and x ions of mg2+

here x is negligible so mg2+ conc will be 0.1

since ksp is given we have

0.1*2x=ksp

you get value of x

now oh- conc is 2x

so ph is 14+log(2x)

now since this is a buffer soln, a small addition of base wont change ph drastically so in 2nd part we can directly say dat answer is C as ph will be slighlty increased due to addn of base... all other options are less than 8.6

for 3rd part im getting 8.6 again... naoh added in 2nd part will react with mgcl2 amd form 0.02 moles mg(oh)2 so total will be 0.12 moles of mg(oh)2 and 0.08 moles mgcl2 remains

now hcl is added and reacts with 0.02 moles of mg(oh)2 and forms 0.02 moles of mgcl2 and total is 0.08+0.02=0.1 moles mgcl2

now we hav 0.1 mgcl2 and 0.1 mg(oh)2
resulting answer will again be 8.6...

sum1 pls correct me if im wrong in 3rd part

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