A student is given an antacid tablet that weighs 5.9200 g. The tablet is crushed and 4.8270 g of the antacid is added to 200. mL of simulated stomach acid. It is allowed to react and then filtered. It is found that 25.00 mL of this partially neutralized stomach acid required 11.7 mL of a NaOH solution to titrate it to a methyl red end point. It takes 27.8 mL of this NaOH solution to neutralize 25.00 mL of the original stomach acid. How much of the stomach acid (in mL) is neutralized by the 4.8270 g crushed sample of the tablet?

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A: To calculate the amount of acetic acid in the vinegar sample, we can use the following balanced…

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A:

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Q: The total volume of 0.25 M H2SO4 used was 27.9 mL. Calculate what volume of 5 M ammonia solution…

A: Molarity of H2SO4 = 0.25 MMolarity of NH3 = 5 M

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Q: To a 25.00 mL volumetric flask, a lab technician adds a 0.450 g sample of a weak monoprotic acid,…

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Q: A student performs a titration and records the following data: Initial buret volume:  11.87 mL…

A: The moles of NaOH used in this titration is = 0.004049 (or) 4.049 × 10-3

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Q: A student is given an antacid tablet that weighs 5.6622 g.  The tablet was crushed and 4.3221 g of…

A:

Q: Consider the titration of 100.0 mL of an unknown concentration of weak monoprotic acid with 0.550 M…

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A:

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Q: calculate

A: We will use gm equivalent of both acid and base

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A:

Q: A student is given an antacid tablet that weighs 5.9320 g. The tablet is crushed and 4.9780 g of the…

A: Given that 29 mL of the solution neutralizes 25 mL of the original solution.  1 mL of the NaOH will…

Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.