1. What would the end point be for a titration of 25.00 mL of0.25 M NaOH with 0.

Question

1. What would the end point be for a titration of 25.00 mL of0.25 M NaOH with 0.10 M HCl?
2. What would the pH be at the end point for thistitration?

So, I started by finding the # moles of NaOH = (.25 M NaOH) (.025L) = .00625 mol NaOH
Then, I found # moles HCl using theoretical 1 L = (.10 M HCL)(1.0L) = .10 mol HCl

I know the limiting reagent is the NaOH. And I know I can usean ICE box to help (I think). And if the chemical equation isNaOH + HCl -> NaCL2 then:

               NaOH              HCl              NaCL
I              .25                    .10               0
C              -x                    -x                 +x
E              .25-x              .10-x              x

So K = [products]/[reactants] = [x]/[.25-x][.10-x] and I am stuckand don't know what to do. Maybe I should be usingM1V1=M2V2? I amconfused, so any help would be appreciated. Thank you.

Explanation / Answer

1) Given data Volume of NaOH, V1 = 25.0 mL    Molarity of NaOH, M1 = 0.25 M Molarity of HCl, M2 = 0.1 M Volume of HCl, V2 = ?    We know that for neutralization reaction M1 V1 = M2V2    V2 =  M1 V1 /M2              =(0.25M)(25.0 mL) /(0.1M)             = 62.5 mL We will get the end point after addition of 62.5 mL ofHCl. 2) Here NaOH and HCl both are strong hence the pH atequivalence point is 7. Because the solution is neutral atequivalence point. We will get the end point after addition of 62.5 mL ofHCl. 2) Here NaOH and HCl both are strong hence the pH atequivalence point is 7. Because the solution is neutral atequivalence point.

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