Question

Titration Lab Sheet Day 2 (ALTERNATE)

Answer 1

Answer: The titration table for the question is given below.

Explanation:

The equation used for the neutralization reaction is:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of an acid.

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base.

Molar ratio is defined as the ratio of the stoichiometric coefficients of of one reactant to the other reactant.

Concentration is the number of moles present in 1 L of solution. Molarity is unit of concentration.

• For Titration 1:

The chemical equation for the reaction of potassium hydroxide and sulfuric acid is given as:

$2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O$

We are given:

$n_1=2\\M_1=?M\\V_1=15mL\\n_2=1\\M_2=0.15M\\V_2=25mL$

Putting values in equation 1, we get:

$2\times M_1\times 15=1\times 0.15\times 25\\M_1=0.125M$

For acid, the molar ratio will be = $\frac{2}{1}$

For base, the molar ratio will be = $\frac{1}{2}$

Concentration of acid $(H_2SO_4$ = 0.125 moles/L

Concentration of base (KOH) = 0.15 moles/L

• For Titration 2:

The chemical equation for the reaction of sodium hydroxide and hydrogen bromide is given as:

$NaOH+HBr\rightarrow NaBr+H_2O$

We are given:

$n_1=1\\M_1=0.25M\\V_1=30mL\\n_2=1\\M_2=?M\\V_2=20mL$

Putting values in equation 1, we get:

$1\times 0.25\times 30=1\times M_2\times 20\\M_2=0.375M$

For acid, the molar ratio will be = $\frac{1}{1}$

For base, the molar ratio will be = $\frac{1}{1}$

Concentration of acid (HBr) = 0.25 moles/L

Concentration of base (NaOH) = 0.375 moles/L

The table for titrations is given below.

Answer 2

#1 is 95L balanced . #2  is 55^3G balanced.