Question

ELIMINARY EXERCISES: Experiment 16 Acid-Base Titrations- Define the following terms associated with titrations: a. standard solutions are base Soluchon When the molanty is alrea known. b. ondpoint is the pant at which color change occurs. C. indicator quie. Watersplutte dues that have one color in basic. 2. Write the balanced formula equation and the net ionic equation for the reaction of sodium hydroxide with hydrochloric acid. Formula equation: Net Ionic equation: 3. What is the molarity of a solution prepared by dissolving 2.24 grams of NaOH in enough deionized water to obtain 500.0 mL of solution? The molar mass of NaOH is 40.0 8. Molarity = 4. What is the color of phenolphthalein indicator in: &. acidic solution b. basic solution 5. Calculate the molarity of acetic acid in a vinear eamnla, knowing that 5.00 mL of vinegar requires 43.50 ml of 0.105 M NaOH to just reach the phenolphthalein endpoint in a titration. Molarity =

Answer 1

1.

Standard solution: A solution of known concentration that is used to determine concentration of other is called standard solution.

End point : This is the neutralization point where acid is completely reacted with base, End point is detected by color change of indicator.

Indicator : Indicator is the chemical substance that changes color depending on the pH. Indicator has different structure in different pH .

Formula equation :

HCl (aq) + NaOH (aq) $\to$ NaCl (aq) + H₂O (l)

Net ionic equation :

H ⁺ (aq) + OH^- (aq) $\to$ H₂O (l)

3.

molarity = $\frac{W_{2}\times 1000}{M_{2}\times V}$

W₂ = mass of NaOH = 2.24 g

M₂ = molar mass of NaOH = 40 g/mol

V = volume of solution = 500.0 mL

then , molarity = $\frac{2.24\times 1000}{40\times 500}$

= 0.112 M

4.

Color of phenolphthalein indicator

acidic solution : Colorless

basic solution: Pink.

5.

balanced reaction is

CH₃COOH + NaOH  $\to$ CH₃COONa + H₂O

moles of NaOH = moles of CH₃COOH

hence

M_acid $\times$ V_acid = M_base$\times$ V_base

or, M_acid$\times$ 5.00 = 0.105 $\times$ 43.50

or, M_{acid =}  $\frac{0.105\times 43.50}{5.00}$

or, M_acid = 0.9135 M

hence molarity of acetic acid in vinegar sample = 0.9135 M