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Purpose (0.5 pt): ha oin the nouede o nouso Standardi2 and Theoretical Backgroun

ID: 589808 • Letter: P

Question

Purpose (0.5 pt): ha oin the nouede o nouso Standardi2 and Theoretical Background (3 pts): Write in complete sentences. Your background should include a balanced chemical e titration and describe the use of indicator and en quat for a general acid/base d point in a titration experiment. Data: Data Table I: Standardization of a NaOH Solution (0.5 pt) Titration #1 #2 #3 Mass of KHP (g) Final Buret Reading (mL) | Initial Buret Reading (mL) | Volume Used (mL) ·We 34·34 3.88 | | | O . I 3G. 7 9 36.66 o. 135 . 8 | 3s .66

Explanation / Answer

Part A) Standardization of NaOH

molar mass of KHP = 204.22 g/mol

moles KHP = moles NaOH = grams KHP/molar mass

molarity NaOH = moles NaOH/volume NaOH

Trial      mass KHP (g)          moles KHP (mol)           Volume NaOH (ml)        molarity NaOH (M)

#1               0.704                       0.00345                              33.88                             0.102                     

#2               0.765                       0.00374                              36.66                             0.102

#3               0.745                       0.00365                              35.66                             0.102

average molarity of NaOH = 0.102 M

Relative Standard Deviation = 0

Part B) Calculated results for unknown acid

moles = molarity x volume

molar mass = grams/moles

Trial      mass unknown acid (g)          Volume NaOH (ml)        moles NaOH (mol)         molar mass (g/mol)

#1                    0.225                                    29.23                              0.003                               75                     

#2                    0.240                                    29.78                            0.00304                              79

#3                    0.224                                    29.11                            0.00297                             75.4

average molar mass = 76.5 g/mol

RSD = sq.rt.[(x - mean)^2/2] = sq.rt.[(2.25 + 6.25 + 1.21)/2] = 2.20

x = individual trial molar masses

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