What is the concentration of NaOH, if the volume of HCl used is 7.62 ml and the concentration of HCl is 0.2 M?
This was a titration experiment. We used 10ml of NaOH and then put 2 drop phenolphthalein. The we poured NaOH drop by drop until it was neutralised.
We know that HCl and NaOH neutralize each other in a 1:1 ratio according to the equation below:
HCl + NaOH --> NaCl + H2O
So if we figure out the number of moles of HCl that were present, we know that this is equal to the number of moles of NaOH at the equivalence point. The concentration of the HCl was 0.2 M and the volume used was 7.62 mL, so multiply the two to get the moles of HCl:
0.2 moles/L * 0.00762 L = 0.001524 moles HCl
This also equals the moles of NaOH at the equivalence point when all of the HCl has been exactly neutralized. Since we have 0.001524 moles of NaOH and you state that the volume of the NaOH was 10 mL, we can divide the two to get the concentration of NaOH:
0.001524 moles/0.010 L = 0.1524 M NaOH