What is the pH of the solution created by combining 1.00 mL of 0.100 M NaOH with
ID: 840743 • Letter: W
Question
What is the pH of the solution created by combining 1.00 mL of 0.100 M NaOH with 8.00 mL of each of the acid solutions? Note: Make simplifying assumptions, do not use the quadratic formula.
-pH after mixing 1.00 mL of 0.100 M NaOH with 8.00 mL of 0.100 M HCl =
-pH after mixing 1.00 mL of 0.100 M NaOH with 8.00 mL of 0.100 M HC2H3O2 =
What are the pH values if you take into account that the 8.00 mL of 0.100 M acid was first diluted with 100 mL water (as it wil be in the experiment you will perform)?
-pH after neutralization of diluted HCl solution =
-pH after neutralization of diluted HC2H3O2 solution =
Explanation / Answer
You will need to calculate the [OH] or [H3O].
HCl + NaOH --> H2O + NaCl
1st determine the number of moles of the acid and the base are available to react. you will need to convert your volumes to liters. 11mL = 0.011L and 8mL = 0.008 L.
moles HCl = 0.008 L x 0.1M = 0.0008 moles HCl
moles NaOH = 0.011L x 0.1 M = 0.0011 moles NaOH
2nd In aqueous solution, the following happens:
NaOH ---> Na+ + OH-
HCl --> H+ + Cl-
So, for every NaOH there is 1 OH and for every HCl there is 1 H. The H and OH react in a 1:1 ratio. Therfore, moles OH = 0.011 and the moles of H are equal to 0.0008.
3rd determine the moles of which is left over after the reaction, OH or H. Since OH is present in the most amount, it will be left over. To determine the moles left over, simply subtract the the moles of H from the moles of OH.
0.0011-0.0008=0.0003 moles OH left over.
4th find the concentration of OH in this solution.
--add the two volumes used 0.011 + 0.008= 0.019 L
--Divide the moles of OH by the volume of the new solution to get the molarity.
0.0003 mol/0.019 L= 0.0158 M [OH}
5th determine pH
pOH =-log [OH]= -log [0.0158 M] = 1.8
14 - pOH =pH
14 - 1.8= 12.2
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