Question

By adding 20 ml 0.1 n hcl to 20 ml 0.001 n koh, the ph of the obtained solution will be

Answer 1

20ml of 0.1NHCl=

0.1

1000

 

×20geq=2×10-3geq.  

20ml of 0.001KOH=

0.001

1000

 

×20geq  

=2×10-5geq.  

∴HCl left unneutralised =2(10-3-10-5)  

=2×10-3(1-0.01)=2×0.99×10-3=1.98×10-3geq.  

Volume of solution =40ml  

∴[HCl]=

1.98×10-3

40

 

×1000M=4.95×10-2  

∴pH=2-log4.95=2-0.7=1.3

Answer 2

The pH of the solution will be 1.31

Explanation:

The relation between normality and molarity is:

$N=n\times M$

where,

N = normality

n = n-factor (acidity for bases and basicity for acids)

M = molarity

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$     .....(1)

• For HCl:

N = 0.1 N

n = 1

So, $M=\frac{0.1}{1}=0.1M$

We are given:

Molarity of HCl solution = 0.1 M

Volume of solution = 20 mL = 0.020 L     (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.1M=\frac{\text{Moles of HCl}}{0.020L}\\\\\text{Moles of HCl}=(0.1mol/L\times 0.020L)=0.002mol$

• For NaOH:

N = 0.001 N

n = 1

So, $M=\frac{0.001}{1}=0.001M$

We are given:

Molarity of NaOH solution = 0.001 M

Volume of solution = 20 mL = 0.020 L

Putting values in equation 1, we get:

$0.001M=\frac{\text{Moles of NaOH}}{0.020L}\\\\\text{Moles of NaOH}=(0.001mol/L\times 0.020L)=0.00002mol$

The chemical equation for the reaction of NaOH with HCl follows:

$HCl+NaOH\rightarrow NaCl+H_2O$

For neutralization:

1 moles of $H^+$ ion will react with 1 mole of $OH^-$ ion

Moles of $H^+$ ion = 0.002 moles

Moles of $OH^-$ ion = 0.00002 moles

Excess moles of $H^+$ ion = (0.002 - 0.00002) = 0.00198 moles

• Concentration of excess acid:

Moles of acid $([H^+])$ = 0.00198 moles

Volume of the solution = [20 + 20] mL = 40 mL = 0.040 L

Putting values in equation 1, we get:

$\text{Molarity of the solution}=\frac{0.00198mol}{0.040L}=0.0495M$

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

$[H^+]=0.0495M$

Putting values in above equation, we get:

$pH=-\log(0.0495)\\\\pH=1.31$

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