Strong Acids and Bases 1800 mL of the stock solution is pipe

Strong Acids and Bases 18.00 mL of the stock solution is pipetted into a 100.00 mL volumetric flask and diluted to the calibration line. 1.50 M stock solution of nitric acid 100 mL volumetric flask When the dilution of nitric acid shown in the diagram above is finished, what is the pH of the final nitric acid solution?

Solution

We know Molarity =moles/volume(in L)

So Moles of Nitric acid in 18ml stock solution= Molarity x Volume(in L)=1.50M x 0.018L (as 18ml=0.018L) =0.027mol

New molarity of nitric acid =moles/volume=0.027mol/0.100L=0.27M

As nitric acid dissociates completely so concentration of H⁺ =[H⁺]= Concentration of HNO₃=0.027M

pH= -log[H⁺]= -log(0.027)= 1.57

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