   Chapter 15, Problem 122MP

Chapter
Section
Textbook Problem

# Consider a solution prepared by mixing the foUowing:50.0 mL of 0.100 M Na3PO4100.0 mL of 0.0500 M KOH200.0 mL of 0.0750 M HCl50.0 mL of 0.150M NaCNDetermine the volume of 0.100 M HNO3 that must be added to this mixture to achieve a final pH value of 7.21.

Interpretation Introduction

Interpretation: The volume of 0.100M HNO3 that must be added to the given mixture to achieve a final pH value of 7.21 is to be calculated.

Concept introduction: The pH value is defined as a figure that is used to express the acidity or basicity of a solution by using the logarithmic scale on which 7 is a neutral. The values lower than 7 shows acidic or greater than 7 shows basic behavior of the solution. Molarity is defined as the number of moles of solute dissolves in a liter of solution.

To determine: The volume of 0.100M HNO3 that must be added to the given mixture to achieve a final pH value of 7.21 .

Explanation

Explanation

Given

The volume and concentration of Na3PO4 is 50.0mL and 0.100M .

The volume and concentration of KOH is 100.0mL and 0.0500M .

The volume and concentration of HCl is 200.0mL and 0.0750M .

The volume and concentration of NaCN is 50.0mL and 0.150M .

The concentration of HNO3 is 0.100M .

The pH value of the mixture is 7.21 .

The conversion of mL to L is done as,

1L=1000mL1mL=103L

Therefore, the conversion of 50mL to L is done as,

1mL=103L50mL=50×103L=0.05L

Similarly, the conversion of 100mL to L is done as,

1mL=103L100mL=100×103L=0.1L

Therefore, the conversion of 200mL to L is done as,

1mL=103L200mL=200×103L=0.02L

The concentration of the mixture is calculated by using the formula,

pH=log[H+]

Substitute the value of pH in the above formula.

pH=log[H+][H+]=10PH=107.21=6.165×108M

The volume (V5) of HNO3 is calculated by using the expression,

M=M1V1+M2V2+M3V3+M4V4+M5V5V1+V2+V3+V4+V5

Where,

• M1andV1 are the concentration and volume of Na3PO4

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