How many milliliters of 1.0 M HCl are needed to exactly neutralize 15 milliliters of 2.0 M Ba(OH)2?

OK, I don't know if this is right but this is my thinking of it. I hope this helps you out.

Let's do problem "a"

Step 1.  Write a balanced equation.

                                       2 HCl  +  Ba(OH)2  >  BaCl2  + 2 HOH  

Step 2.  Underneath the HCl and Ba(OH)2 in the equation write what you are given and what you need to find out.

                                    2 HCl        +       Ba(OH)2  >  BaCl2  + 2 HOH

                                 C = 0.130 mol/L    v = 56 mL

                                  V = ?                   C = 0.109 mol/L

Step 3.  They ALWAYS provide a way to calculate the number of moles of one of the substances.  In this case it is Ba(OH)2 because you have C and V

                         n = C X V            n = 0.109 x 56 = 6.104 millimoles

Step 4.  Use the equation to figure out how many moles of the unknown (HCl) you will need.

               the equation says you need TWICE as many moles as you have of Ba(OH)2.  So you will need 6.104 millimoles x 2 or 12.208 milimoles.

Step 5.  Now that yo have moles and concentration of HCl you can now calculate volume

                  C = n / V, so V = n / C

                                 12.208 millimoles / 0.130 mol/L = 93.9 mL

Use this method for all these kind of problems.  

In B) you will have to convert he g of NaOH into moles by 0.240 / 40.00 g/mol and then following trhe rest of the procedure.

C) is done the same way.  convert the g of Na2SO4 into moles first.

d) is almost the same as a) except this time when you find the moles of Ca(OH)2 you just have to convert the moles into mass by multiplying moles x Molar mass of Ca(OH)2.

Best of luck.

Explanation:

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