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Sonja [21]
3 years ago
15

10) A student’s calculation w as found to have a 15.6% error, and the actual value w as determined to be 25.7 mL. W hat are the

tw o possible values for the student’s experimental measurement?
Chemistry
1 answer:
goblinko [34]3 years ago
7 0

Answer:

The actual value = 25.7 ml

The error = 15.6%

This means that the value of error = 0.156 * 25.7 = 4.0092 ml

Now, the error percentage found means that the student got the value either greater than the actual one with the value of error or less than the actual one with the value of error.

This means that the two possible readings are:

either : 25.7 + 4.0092  = 29.7092 ml

or : 25.7 - 4.0092  = 21.6908 ml

Explanation:

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You have 500.0 ml of a buffer solution containing 0.30 m acetic acid (ch3cooh) and 0.20 m sodium acetate (ch3coona). what will t
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First, we should get moles acetic acid = molarity * volume

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                                                           = 0.2 M * 0.5L

                                                           = 0.1 mol

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moles acetic acid after adding OH- = (0.15-0.02) 
                                             
                                                            =  0.13M                                       

moles acetate after adding OH- =  (0.1 + 0.02)

                                                      =   0.12 M

Total volume = 0.5 L + 0.02 L= 0.52 L

∴[acetic acid] = moles acetic acid after adding OH- / total volume

                        = 0.13mol / 0.52L

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and [acetate ) = 0.12 mol / 0.52L
 
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by using H-H equation we can get PH:

PH = Pka + ㏒[salt/acid]

when we have Ka = 1.8 x 10^-5

∴Pka = -㏒Ka 

        = -㏒ 1.8 x 10^-5

       = 4.7

So by substitution:

∴ PH = 4.7 + ㏒[acetate/acetic acid]

         = 4.7 + ㏒(0.23/0.25)

        = 4.66
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