2. A 3.00 M NaoH solution was prepared by your instructor to be used in this exp
ID: 476731 • Letter: 2
Question
2. A 3.00 M NaoH solution was prepared by your instructor to be used in this experiment. If you wanted to make 2.00 L of a 3.00 M NaoH solution, how many grams of solid NaoH pellets would you need to dissolve in water? grams 3. You added 10.0 mL of 3.00 M NaoH into your reaction flask. How many equivalents of NaoH did you add to your reaction flask as compared to methyl benzoate? If you do not know what equivalents are, look in the laboratory manual appendix. equivalents 4. You quench your reaction by adding concentrated hydrochloric acid. Concentrated HCI is 12.0 M. How many mL of concentrated HCl would be required to completely neutralize the 10.0 mL of 3.00 M NaoH that you initially added to your reaction flask? Completely neutralize means to get to a pH of 7 (make a neutral pH). mLExplanation / Answer
2) mol of NaOH = 3 mol/L * 2 L = 6 mol
molecular weight of NaOH = 39.997 gm/mol
grams of solid NaOH pellets required = 39.997 gm/mol * 6 mol = 239.982 gms
3) mol of NaOH = (10/1000)L * 3 mol/L = 0.03 mol
equivalent of NaOH = 0.03mol* 1 eq = 0.03 equivalents
4) PH = PKa + log [ Cl-/ HCl ]
7 = -7 + log [ Cl-/ HCl ]
100 = Cl-/ HCl
[Cl-/ HCl ] = 1
[Cl-/ HCl ] = [ 0.03-X / X ]
1 = 0.03-X / X
X = 0.015 mol = 15 mmoles of HCl is required
Multiply this with molarity of HCl = 15 mmol * (1 mL/ 12*10-3 mol) = 1250 mL of HCl is required
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