Titrations worksheet Titrations worksheet 1) It takes 83 ml. of a 0.45 M NaOH so

Question

Titrations worksheet Titrations worksheet 1) It takes 83 ml. of a 0.45 M NaOH solution to neutralize 235 ml. of an HCl solution. What is the concentration of the HCl solution? 2) You are titrating an acid into a base to determine the concentration of the base. The endpoint of the neutralization is reached but the stopcock on the buret sticks slightly and allows a few more drops of acid to fall into the solution. How will this affect your calculations for the concentration of the base? 3) It takes 38 mL of 0.75 M NaOH solution to completely neutralize 155 mL of a sulfuric acid solution (H2SO4). What is the concentration of the H2S04 solution? 4) A few small drops of water are left in a buret that is then used to titrate a base into an acid solution to determine the concentration of the acid. Will this small amount of water have any effect on the determined value for the concentration of the acid? If so, how is it affected? 5) It takes 12.5 mL of a 0.30 M HCI solution to neutralize 285 mL of NaOH solution. What is the concentration of the NaOH solution? 6) Lulu Labwrecker carefully pipets 25.0 mL. of 0.525 M NaOH into a test tube. She places the test tube into a small beaker to keep it from spilling and then pipets 75.0 ml of 0.355 M HC1 into another test tube. When Lulu reaches to put this test tube of acid into the beaker along with test tube of base she accidentally knocks the test tubes together hard enough to break them and their respective contents combine in the bottom of the beaker. Is the solution formed from the contents of the two test tubes acidic or basic? What is the pH of the resulting solution?

Explanation / Answer

Ans 1 :

The reaction is given as :

NaOH + HCl = NaCl + H2O

To calculate the concentration of HCl , we can use the formula , M1V1 = M2V2

where M is molarity and V is volume

So putting in all the values we get :

0.45 x 83 = M2 x 235

M2 = 0.159 M

So the concentration of HCl solution will be 0.159 M.

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