PH at the equivalence point?

Ksp and Buffers question: finding pH at equivalence point?

• Answer:

  well we need to look at two specific equations on is the reaction of the acid and the base HNO2+NaOH<---->NaNO2 +H2O now since we dont directly have any OH- or H+ we can assume that nitrous acid is a weak acid and NaNO2 is its conjugate base thus we need to treat it as a buffer so now we look at the moles of each (normally I would use the ICE table for this but It is really hard to do it on here so) since we have the same molarity we know that it will take 25mL of NaOH to reach equivalence point now we use the ice table to see how much of our NaNO2 is left over after reaching the equivelence point. so we take the number of moles of HNO2 and since it is a one to one ratio we know that that is how many moles of NaNO2 we will have before the titration so now we just add the moles of NaOH used which will give us zero moles of NaOH and zero moles of HNO2 and 7.50E-3 mole of NaNO2 Now that we know the moles of NaNO2 we can set up the hydrolosis equation for the salt NaNO2 NaNO2(aq)---->Na+(aq) +NO2- since Na+ is the conjugate of a strong base we know that it will not hydrolyze only the NO2- will so now we set up the hydrolosis equation NO2-(aq)+H2O(l)<---->HNO2 +OH- the OH tells us that we have a basic solution thus we need the Kb to set up the equalibrium equation Kb=Kw/Ka thus Kb=1.0E-14/4.50E-4 Kb=2.22E-11 we now need the molarity of the NO2- from the titration remember we added 25mL to the original solution mole NO2-=7.50E-3 mL=50.00 M NO2-=(7.50E-3 mole)/(0.050L) [NO2]=1.5E-1M now using the ICE table we come up with the following equation Kb=[HNO2][OH-]/[NO2-] and since the Kb value is so low we can ignore the x on the bottom leaving us with the follwing equation 2.22E-11=(X*X)/0.150M multiply the .150 out and take the sqrt Sqrt(2.22E-11*0.150)=X X=1.83E-6 mole since you know that [OH-]=X you can just take the negative log of the x value to come up with pOH=5.74 subtract this from fourteen to find your pH value pH=14.00-5,74 pH=8.26 My answer is a little off because I didn't round until the very end but it is still correct