   Chapter 15, Problem 71QAP ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
210 views

# 71. A sample of sodium hydrogen carbonate solid weighing 0.1015 g requires 47.21 ml. of a hydrochloric acid solution to react completely. HCl ( aq ) + NaHCO 3 ( s ) → NaCl ( aq ) + H 2 O ( l ) + CO 2 ( g ) Calculate the molarity of the hydrochloric acid solution.

Interpretation Introduction

Interpretation:

The molarity of the hydrochloric acid solution is to be calculated.

Concept Introduction:

Acid is a substance that donates H+ ions, whereas a base is the substance which accepts H+ ions. A chemical reaction that occurs between acid and base is known as neutralization reaction. In the neutralization reaction, the OH ions of the base combines with H+ ions of an acid, which results in the formation of water as a side product.

Molarity is used to find out the concentration of solution.

The molarity is calculated by the formula,

Molarity=NumberofmolesofsoluteVolumeofsolutionL.

Explanation

The mass of sodium hydrogen carbonate NaHCO3 is given to be 0.1015g and the volume of HCl solution is given to be 47.21mL.

The molar mass of NaHCO3 is 84.00g/mol.

The number of moles of NaHCO3 is calculated by the formula,

Numberofmoles=MassgMolarmass        (1)

Substitute the values of molarity and volume of NaHCO3 in the equation (1).

Numberofmoles=0.1015g84.00g/mol=0.00121moles

The chemical equation for the given reaction is shown below.

HClaq+NaHCO3sNaClaq+H2Ol+CO2g

The above equation indicates that both the reactants react in the ration of 1:1

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