We performed an experiment of a reaction between NaOH, a strong base, and HCl, a strong acid. However, we were given 1 M HCl and 0.3 M NaOH.  In order to demonstrate this reaction, we wanted our two reactants to be of the same molarity and volume.  We set our goal at: 0.3 M and 100 mL.

In order to reach two reactions of the same molarity, we used the formula M1V1=M2V2 to find out how much water was needed to dilute the 1 M HCl and reach 100 mL of 0.3 M HCl.  Plugging in the numbers into the formula we got: (100 mL HCl)(0.3 M HCl) = (30 mL HCl)(1 M HCl).  According to the formula, we needed to add 30 mL of 1 M HCl and 70 mL of water in order to make 100 mL of 0.3 M HCl.  (Then, we would be able to use the reactants: 0.3 M HCl and 0.3 M NaOH.)  In order to do this, we used two 100 mL graduated cylinders. 

We then measured out 30 mL of 1 M HCl and 70 mL of water.

(NOTE: TO PERFORM THIS EXPERIMENT, WE DONNED GOGGLES AND CLOSE-TOED SHOES FOR MAXIMUM SAFETY.) 
 

Pouring the 70 mL of water into the 1 M HCl to create 100 mL of 0.3 M HCl.  

Next, to begin our experiment, we used phenolphthalein to indicate the pH level of the solution.  As defined in our glossary, it is an indicator that turns pink when the pH is above 8.  We then measured out 100 mL of 0.3 M NaOH and then added drops of phenolphthalein to the solution until it turned pink. 

We then added the 100 mL of 0.3 M HCl to the 100 mL solution of 0.3 M NaOH.