1. Obtain 10.0 mL of 0.10 M HCl. Transfer 9.0 mL to a 50 mL or 100 mL beaker lab
ID: 820643 • Letter: 1
Question
1. Obtain 10.0 mL of 0.10 M HCl. Transfer 9.0 mL to a 50 mL or 100 mL beaker labeled HCl #1. Dilute 1.0 mL of the 0.10 M HCl to 10.0 mL with deionized water in your graduated cylinder, transfer to a 50 mL or 100 mL beaker, and label HCl #2. Measure the pH of each solution.
2. Obtain 10.0 mL of 0.10 M HC2H3O2. Transfer 9.0 mL to a 50 mL or 100 mL beaker labeled HC2H3O2 #1. Dilute 1.0 mL of the 0.10 M HC2H3O2 to 10.0 mL with deionized water in your graduated cylinder, transfer to a 50 mL or 100 mL beaker, and label HC2H3O2 #2. Measure the pH of each solution.
Now, calculate the (1) molarity and (2) calculated pH for each of the following:
HCl #1 (measured pH = 1.50)
HCl #2 (measured pH = 2.50)
HC2H3O2 #1 (measured pH = 3.00)
HC2H3O2 #2 (measured pH = 3.50)
Explanation / Answer
Hcl
1. moles of hcl in 9 ml = 0.1*0.009 = 0.9milli moles
conc=[H] =0.1 M
molarity=0.1M
PH=-log[H]= -(-1)=1
2.moles of hcl = 0.1*0.001= 0.1 millimoles
con, [H]= 0.1/10 = 0.01
molarity=0.01M
PH=-log[H]=2
HC2H3O2
1. conc ,[OH]= 0.1M
molarity=0.1M
[OH][H]=10^-14 ...[H]=10^-13
PH=-log[H]=13
2.moles =0.1 millimoles
conc=[OH]=0.1/10=0.01M
molarity=0.01M
[OH]*[H]=10^-14.......[H}=10^-12
PH= 12
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