## 2012 Final Q7b

Sydney Wu 2M
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### 2012 Final Q7b

"Calculate the pH at the stoichiometric point for the titration of 15.00mL of 0.100M HCOOH (Ka = 1.8 x 10-4) with 0.150M NaOH."

Can someone please explain how to start this? I'm looking at the answer key and lecture notes, but I am completely lost on the logic of how to start and proceed with the calculations (once I get to the ICE box, I'll know what to do, but I just don't know how to get there...).

Thank you!

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### Re: 2012 Final Q7b

Since this question is the titration of a strong base with a weak acid your first step is to convert the given Ka into Kb using Ka*Kb=Kw. This is because at the stoichiometric point the HCOOH has broken into its conjugate base and exists as the salt $HCOO^{-} Na^{+}$. This state will be the reactant side on your ICE chart equation and the product side will contain your now neutralized HCOOH and $OH^{-}$. To fill in the ICE chart you must determine the concentration of $HCOO^{-}$ in solution. You can determine the number of moles of HCOOH that was present to begin with by multiplying the liters times the concentration of the HCOOH.

mol HCOOH=0.015L * 0.100 M HCOOH= $1.5*10^{-3}$.

You know that the moles of $OH^{-}$ must be equal to the moles of HCOOH and the moles of $HCOO^{-}$ at the stoichiometric point.

Using the moles of NaOH you can solve for the volume of NaOH added, then find the total volume, then use the total volume of solution to calculate the M of the $HCOO^{-}$ for your ICE chart.

Vol NaOH= $1.5*10^{-3}$/(0.150 M NaOH)= 10.0 mL. Total Volume= 10mL + 15mL = 25mL.
Molarity $HCOO^{-}$ = ($1.5*10^{-3}$)/(0.025L)= 0.06 M. <-----This is what you put in your ICE chart for I of $HCOO^{-}$. And if you got the ICE chart down the rest is just calculating the pH.