I don't see any number given for the molarity of HNO2 initially. From your work I think you've used 0.015 M for HNO2 but the way I read the problem that 0.015 M is the concentration of the Ca(OH)2. I don't think you can determine the equivalence point without knowing the concentration of the HNO2. Let me pick a convenient number and you can follow my calculations with whatever the correct value is in your problem. I will use 0.015M for HNO2 so the calculation will be simplified.
Ca(OH)2 + 2HNO2 ==> Ca(NO2)2 + 2H2O
I like to work in millimoles. So 20.00 mL of 0.015 M HNO2 = 20.00 x 0.015 = 0.300 millimoles HNO2 initially. So we must have
Convert to millimoles Ca(OH)2 = 0.300 mmols HNO2 x (1 mols Ca(OH)2/2 mols HNO2) = 0.15.
Since M = mmols/mL then 0.15 = 0.015/mL so mL Ca(OH)2 = 0.150/015 = 10 mL. (The fact you arrived at the same number is a coincidence because I picked 0.015 M for the HNO2). So + 2 mL after tje eqivalence point will be 12 ml of 0.015 M Ca(OH)2. What do we have at this point? We have zero HNO2 and zero Ca(OH)2 + 2 mL x 0.015 M Ca(OH)2 [that's 0.03 mmol) and we have a volume of 20 mL + 10 mL + 2 mL = 32 mL total. Also we have 10 mmols Ca(NO2)2. The new concentration of the Ca(OH)2 after the eq point is M = mmols/mL = (2 x 0.015)/32 mL = ?. The (OH^-) will be twice that and you can calculate the pH from that. Frankly, I think such a low concentration of Ca(OH)2 after the eq point means we may have needed to take into consideration the hydrolysis of the Ca(NO2)2 salt but I didn't go into that. My assumption is that this is a beginning chemistry class and that would not be included in a beginners class. Good luck.
Determine the ph when 2.00ml of additional Ca(OH)2 is added after the equivalence point.
Started with 20.00ml of HNO2 and it is titrated with 1.50M Ca(OH)2.
Ka= 4.0 x 10^-4
So I found the equivalence point = 0.02L x 1.50M HNO2/ 3.00M (cuz 2 mols of OH so 1.5 x2)= 0.01L
So i know we have to find moles of both HNO3 and Ca(OH)2.
moles of HNO3 = 0.03 mol
I'm having trouble finding the moles for Ca(OH)2 . Since the question said after an additional 2ml after equivalence point the volume will be (eq volume + 2ml) = 0.012L? So when finding moles of Ca(OH)2 do I do (0.012x3.00M)?
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