(10pts) Post-Lab Questions Answer the following questions. (2pts) Suppose the concentration of the NaOH solution was 0.5 M instead of 0.1 M. Would this titration have required more, less, or the same amount of NaOH solution for a complete reaction? Choose... (2pts) Suppose you used 0.5 M NaOH to titrate your vinegar sample instead of 0.1 M. What effect does the concentration of base added have on the reliability of the results of your titration? Choose... (2pts) Consider a 0.238 M aqueous solution of sodium hydroxide, NaOH. How many grams of NaOH are dissolved in 23.46 mL? (2pts) How many individual hydroxide ions (OH−) are found in 23.46 mL? (2pts) How many moles of sulfuric acid, H2SO4, are neutralized by 23.46 mL of 0.238 M NaOH(aq)? [Hint: begin by writing a balanced equation for this neutralization reaction.] Total: -/28 pts

A) Since the number of moles of acetic acid in the vinegar sample is constant, the number of moles of NaOH reacting with acetic acid is constant as well.

Volume = number of moles / concentration

Assuming, number of moles = 1, concentration = 0.1 M

Volume = 1/0.1 = 10

With increase in concentration,

Volume = 1/0.5 = 2

Therefore, less amount of NaOH will be required

B) An increase in concentration of base added will results in lesser volume of based used. This will result in an increase in percentage error as well as decrease innthe accuracy of determining the endpoint. Hence, reliability decreases

C) mass = concentration × volume × molar mass

Molar mass of NaOH = 40.0 g, volume of NaOH = 23.46 mL = 0.02346 L, concentration = 0.238 M

Mass of NaOH = 0.238 × 0.02346 × 40

Mass of NaOH = 0.223 g

D) number of moles of NaOH in 0.223 g = 0.233/40 = 0.00558 moles

1 mole of NaOH produces 1 mole of hydroxide ions; 0.00558 moles of NaOH will produce 0.00558 moles of hydroxide ions

Number of hydroxide ions in 0.00558 moles = 0.00558 × 6.02 × 10²³

Number of hydroxide ions = 3.36 × 102²¹

E) H₂SO₄ + 2 NaOH ----> Na₂SO₄ + 2 H₂O

1 mole of H₂SO₄ requires 2 moles of NaOH for complete neutralization

There are 0.00558 moles of NaOH in 23.46 mL of 0.238 M NaOH solution.

Moles of H₂SO₄ required to neutralize 0.00558 moles of NaOH = 0.00558/2

Moles of H₂SO₄ neutralized = 0.00279 moles of H₂SO₄