1. A local company sent you their green alternative for window cleaner to be tested for percent (w/v) acetic acid content. For your experiment, you first standardized your NaOH titrant with 0.8053 g of (99.80 % purity) KHP. You used 40.60 mL of NaOH for your standardization. After that you then analyzed a 10.00 mL sample and found that you needed 43.20 mL NaOH to reach the end point.

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1. A local company sent you their green alternative for window cleaner to be tested for
percent (w/v) acetic acid content. For your experiment, you first standardized your
NaOH titrant with 0.8053 g of (99.80 % purity) KHP. You used 40.60 mL of NaOH
for your standardization. After that you then analyzed a 10.00 mL sample and found
that you needed 43.20 mL NAOH to reach the end point.
Summary of results:
Sample analysis
Standardization
0.8053 g
KHP Weight (g)
Purity
NaOH (mL) used
Determine the following:
a. Molarity of NaOH
b. % (w/v) acetic acid
Volume of sample
NaOH (mL) used
50.00 mL
99.80%
33.20 mL
40.60 mL
Transcribed Image Text:1. A local company sent you their green alternative for window cleaner to be tested for percent (w/v) acetic acid content. For your experiment, you first standardized your NaOH titrant with 0.8053 g of (99.80 % purity) KHP. You used 40.60 mL of NaOH for your standardization. After that you then analyzed a 10.00 mL sample and found that you needed 43.20 mL NAOH to reach the end point. Summary of results: Sample analysis Standardization 0.8053 g KHP Weight (g) Purity NaOH (mL) used Determine the following: a. Molarity of NaOH b. % (w/v) acetic acid Volume of sample NaOH (mL) used 50.00 mL 99.80% 33.20 mL 40.60 mL
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