Question

4. (3 pts) If 2.55 g of NaOH were added to a 100.0 mL solution containing 0.10 M Ca(NO3)2, what (Ca?') remains in solution? Is there complete precipitation of Ca(OH)2?

Answer 1

Volume of solution - 100.ome 01L Malarity of solution of Calmus), = 0.101 Malarity of solution : no. of males of solute Total volume (in L) 1. 0.1 = no. of males of (a(NO3)2 Now 8.559 no. of meles of La CNO3)2 = 0.01 adoled amount of Naut's Molar mou of Naona Molar mau of nat Melar may S of 'o' + molar man ofn = 23 +18+1 = tog mel no. of meles of Naon = wt. of Naon = 8.55g Molar mali of Naon 40g/mel : = 0.06375 mole Reaction occured here is Ca(NO2) + 2NaOH (ALOH) + 2NaNO3 + Hoon 0.0i mele 0.06375 mele Here 2 make of Naon react with one mele of Ca(NO? to form one mele of Calone and g mele of Nanio So as Naon (0.06375 mele) is present in higher amount as compared to (a (NO3)2 Cool mele) All Ca(NO3)2 will react Naon. to produce ool mole of Caloh₂ and 2xool mole of Na(NO) so nown deft after reaction with Carnos= 0.06375 -0.02 as 2 mele of Naon react with = 0.04315 one mole of Ca(NO3h thats. 0.02 is subtracted from 0106375.

SU 1. NGON left so conch of 0.04375 meles, voluene - pol 012 Conds 0.04375 = 0.4315M OTOT O.IL And (a cong = ca 21+ 2010 ko = 6x156 for ksp= [(a 2+ ] [onoga lacon oxióba [1a2+] x 0.4375*0.4375 / [ca2+] = 3.13x105m) soon there will no complete As co2t is present in precipitation of Ca24.