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Titration Lab Sheet Day 2 (ALTERNATE)

Titration Lab Sheet Day 2 (ALTERNATE)-example-1
User Jon Hess
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2 Answers

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#1 is 95L balanced . #2  is 55^3G balanced.
User Gabriel Mesquita
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Answer: The titration table for the question is given below.

Step-by-step 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* M_1* 15=1* 0.15* 25\\M_1=0.125M

For acid, the molar ratio will be =
(2)/(1)

For base, the molar ratio will be =
(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* 0.25* 30=1* M_2* 20\\M_2=0.375M

For acid, the molar ratio will be =
(1)/(1)

For base, the molar ratio will be =
(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.

Titration Lab Sheet Day 2 (ALTERNATE)-example-1
User Gadolf
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