The following are some calculation questions related to the experiment. Please i

Question

The following are some calculation questions related to the experiment. Please include the questions and their answers in your report:

a) A 0.05 L solution of 0.5 mol/L NaOH was titrated until neutralized into a 0.025 L sample of HCl. What was the concentration of the HCl?

b) 50 L of 2.0 mol/L Hydrochloric acid is titrated with sodium hydroxide to form water and sodium chloride. How many moles of sodium hydroxide are consumed in this reaction?

c) 100 L of 5 mol/L NaOH are required to fully titrate a 50 L solution of HCl. What is the initial concentration of the acid?

Explanation / Answer

The following are some calculation questions related to the experiment. Please include the questions and their answers in your report:

a) A 0.05 L solution of 0.5 mol/L NaOH was titrated until neutralized into a 0.025 L sample of HCl. What was the concentration of the HCl?

The reaction of HCl and NaOH arer as follows:

HCl + NaOH = NaCl + H2O

Moles of NaOH= Molarity * volumes in L

= 0.5 moles /L* 0.05L

= 0.025 MolesNaOH

The reaction occurs in 1:1 ratio thus the number of moles of acid and base are equal.

Number of acid = 0.025 Moles HCl

Molarity or concentration = number of moles / volume in L

= 0.025 Moles HCl /0.025 L

= 1.0 M HCl

b) 50 L of 2.0 mol/L Hydrochloric acid is titrated with sodium hydroxide to form water and sodium chloride. How many moles of sodium hydroxide are consumed in this reaction?

The reaction of HCl and NaOH arer as follows:

HCl + NaOH = NaCl + H2O

Moles of HCl= Molarity * volumes in L

= 2.0 moles /L* 50 L

= 100 Moles HCl

The reaction occurs in 1:1 ratio thus the number of moles of acid and base are equal.

Number of acid =Number of Base; NaOH = 100 Moles

c) 100 L of 5 mol/L NaOH are required to fully titrate a 50 L solution of HCl. What is the initial concentration of the acid?

The reaction of HCl and NaOH arer as follows:

HCl + NaOH = NaCl + H2O

Moles of NaOH= Molarity * volumes in L

= 5 moles /L* 100 L

= 500 Moles NaOH

The reaction occurs in 1:1 ratio thus the number of moles of acid and base are equal.

Number of acid = 500 Moles HCl

Molarity or concentration = number of moles / volume in L

= 500 Moles HCl /50 L

= 10 M HCl

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