ACID-BASE NEUTRALIZATION REACTIONS. In this lab, we were given several vials whi

Question

ACID-BASE NEUTRALIZATION REACTIONS. In this lab, we were given several vials which each contained NaOH solutions at various pH levels. Vial 1 had a pH of 9. We're working with the chemical equation: HCl + NaOH <--> H2O + NaCl, with a Net Ionic equation of H+ + OH- = H2O

Q) Calculate the concentration of OH- of the solution in vial 1 at room temperature. Given that you placed 5 mL of this solution into each vial, how many moles of OH- are in each vial before any additional solution is added?

I found the [OH-] in the following manner:
pOH = 14 - pH // 14 - 9 = 5
[OH-] = 10-pOH // 10-5 = 1 x 10-5 M = [OH-]

I am confused as to how to find the number of moles. Do I use the [OH-]? Is it a stoichiometric thing?

Explanation / Answer

Actually , the concentration you have calculated is nothing but the molarity of the solution .

And we know that molarity is equal to the no. Of moles divided by the volume of solution in litres.

So knowing the molarity and the volume , we can calculate the no. Of moles.

So V=5ml given or 5/1000 litres

And molarity is 10^-5

So n= M×V

n= 10^-5 × (5×10^-3)

n= 5×10^-8 is the ans.

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