ACID-BASE TITRATION PRELAB QUESTIONS NAME niele Piver 1. A 4.00 mL sample of vinegar (active ingredient, acetic acid) was titrated with 0.1250M NAOH according to the following equation: HC2H3O2 NaOH NaC2H3O2 + HOH 25.19 mL of base was needed to reach the end point. Calculate the molarity of the acetic acid. 2.molt2Ha02 Sas1qL NaoH O.1a5om mosaoH basH tmoi lcatt 20 1 molaott 1 L o031 4825mo HCaHaa O04004H0zte2 Calculate the grams of acetic acid in one liter of the vinegar. 2mol tal20 o.18 CaH302 2. O.78 7M of vinegar in The 4.00 mL sample of vinegar had a mass of 4.056 g. Calculate the one liter. grams 3. Calculate the % acetic acid in the vinegar. 4. g acetic acid x 100 g vinegar % acetic acid 95

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Consider the following operations on the number 8.83 x 10-². Without using a calculator, decide which would give a significantly smaller value than 8.83 x 10-2, which would give a significantly larger value, or which would give essentially the same value. 8.83 x 10-2 +3.23 x 105 (same 8.83 x 10-2-3.23 x 105 smaller 8.83 x 10 2 x 3.23 x 105 larger v 8.83 x 10-2/3.23 x 105 smaller

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» Significant Figure Rules 01 >> 02 >> 03 >> 04 >> 05 >> All nonzero digits are significant. Zeros that appear between other nonzero digits are always significant. Zeros that appear in front of all of the nonzero digits are called left-end zeros. Left-end zeros are never significant. Zeros that appear after all nonzero digits are called right-end zeros. Right-end zeros in a number that locks a decimal point are not significant. ▬▬▬ 美銀銀時 ☐☐ Right-end zeros in a number with a decimal point are significant. This is true whether the zeros occur before or ofter the decimal point. 237 has three significant figures. 1.897 has four significant figures. 39,004 has five significant figures. 5.02 has three significant figures. 0.008 has one significant figure. 0.000416 has three significant figures. 140 has two significant figures. 75,210 has four significant figures. 620.0 has four significant figures. 19.000 has five significant figures. For multiplication and division problems, the answer…