What is the pH of the solution after the addition of 84.0 mL of 0.100 M HNO3?

Question

In this question, the original solution we are referring to is a 0.0100 mol sample of solid Cd(OH)2 with a Ksp of 5.3 x 10^(-35)

I've found the molar solubility of Cd(OH)2 in pure water to be 1.1x10^(-5) and to have a pH of 9.35

I know 84mL of a 0.1 HNO3 solution yields 0.0084 moles of HNO3 and I think I would be using some sort of ICE table but I'm not sure of the equation.

Since HNO3 (Nitric Acid) is a strong acid then it should dissociate completely when it hits the water so we're probably only concerned with the [H ] ions and how they will react with the Cd(OH)2 solution.

I see Cd(OH)2 breaking up into Cd 2 and 2(OH)-

so I would think two of the H ions will combine with the 2(OH)- to form H20.

We have more Cd(OH)2 than H ions so all the the H will get used up to form water.

since we have .01 moles of Cd(OH)2 and each of those moles produce 2 OH- ions then we'll have 0.02 moles of the hydroxy ions (is hydroxy the right word???)

So, I'll take the .02 moles OH- and subtract the .0084 moles of H which yields 0.0116 moles of OH,

to obtain [OH-] I'll take 0.0116 and divide by our new total volume which is now 184mL and

0.0116/0.184 = 0.063 M

then I'll take [H3O ] = Kw/[OH-] = 10^(-14)/.063 = 1.58e-13

then pH = -log(1.58e-13) = 12.80

BUT that doesn't make any sense because we started w/ a pH of 9.35 and adding nitric acid should lower the pH.

What am I missing?

A follow up question to this is how many milliliters of 0.100 M HNO3 must be added to completely neutralize the Cd(OH)2 but, it seems my thinking is already flawed somewhere so I need to get things straight. I would think, 0.02 moles of HNO3 would do it (simply convert to mL) but am not sure

Answer 1

based upon the molar solubility which you found to be 1.1x10^(-5), ... & the Ksp for Cd(OH)2 in my text,.... you got a typo in your Ksp,...

your Ksp should be 5.3e-15,.. to give that Molar solubility

===========================

you know 84mL of a 0.1 HNO3 solution yields 0.0084 moles of HNO3

you know you have .01 moles of Cd(OH)2

You know that 2 moles of HNO3 neutralize 1 mole of Cd(OH)2, so 0.0084 moles of HNO3 wil neutralize 0.0042 moles of Cd(OH)2 releasing 0.0042 moles of Cd+2  

now 0.0042 moles of Cd+2 in 84 ml , gives a molarity...:

0.0042 moles / 0.084 litres = 0.05 Molar Cd+2

the resultant pH depends upon the amount of [OH-] that Cd(OH)2 will still release into a 0.05 Molar Cd+2 solution:

Ksp of 5.3 x 10^(-15)

5.3 e-15 = [Cd] [OH]^2

5.3 e-15 = [0.05] [OH]^2

[OH]^2 = 1.06e-13

[OH-] = 3.26 e-7

pOH = 6.49

14 - pOH = pH

your answer is 7.51

====================

adding through an edit :

if you had added an excess of HNO3, you would expect an acidic pH.  but the addied HNO3 was converted into H2O & (NO3)-1 ions

you are absolutely right about: "0.02 moles of HNO3 would do..." neutralize the 0.01 moles of Cd(OH)2.

& right about: ..."simply convert to mL..." :

0.02 moles @ 0.1 mol / litre = 0.2 litres

Your last answer is : 200ml of 0.1 M HNO3 will completely neutralize the Cd(OH)2

the flaw, was assuming that the residual Cd(OH)2 would release all of its un-neutralized OH- , but  it couldn't do that,... it had to maintain the Ksp