   Chapter 15, Problem 94AP ### 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
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# 94. Baking soda (sodium hydrogen carbonate. NaHCO3) is often used to neutralize spills of acids on the benchtop in the laboratory. What mass of NaHCO3 would be needed to neutralize a spill consisting of 25.2 mL of 6.01 M hydrochloric acid solution?

Interpretation Introduction

Interpretation:

The mass of NaHCO3 that would be needed to neutralize a spill consisting of 25.2mL of 6.01M

HCl solution.

Concept Introduction:

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. During neutralization reaction, water is released as a side product.

Molarity is used to determine the concentration of the substance in a solution.

Explanation

The volume and molarity of HCl solution is given to be 25.2mL and 6.01M respectively.

The conversion of units of 25.2mL into L is done as,

25.2mL=25.21000L=0.0252L

The number of moles of HCl solution is calculated by the formula,

NumberofMoles=Volume×Molarity        (1)

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

NumberofMoles=0.0252L×6.01M=0.151moles

The balanced equation of HCl and NaHCO3 is shown belwo.

HClaq+NaHCO3sNaClaq+H2Ol+CO2g

The above equation implies that one equivalent of HCl reacts with one equivalent of NaHCO3.

Hence, the number of moles of NaHCO3 required to neutralize HCl is calculated by the formula,

MolesofNaHCO3required=MolesofHCl×1molNaHCO31molHCl        (2)

Substitute the value of moles of HCl in the equation (2)

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