A flask containing 14.92mL of nitric acid solution was titrated with 0.135M NaOH

Question

  
A flask containing 14.92mL of nitric acid solution was titrated with 0.135M NaOH. 16.93mL of the base were required to reach the endpoint. What is the concentration of the nitric acid solution?

For the first part I used m1v1 = m2v2 and got 0.119M. I don't know if is right or not.

The nitric acid solutionhas a density of 1.09g/mL. Using this density and the molarity you calculated above, fill in the table. (please show all work so I can see how to do it).

I don't know what volume of original solution is. 14.92mL or 16.93mL?

the chart needs the volume of original nitric acid solution ___________

nitric acid in grams ________ and moles_____

water grams ________

total solution _______

1, a. A flask containing 14 92 mL of a nmitric acid solution was titrated with 0:135 M NaOH. 16 93 mL. of the base were required to reach the endpoint. What is the concentration of the nitric acid solution? malarity M molarity: . I 19

Explanation / Answer

A flask containing 14.92mL of nitric acid solution was titrated with 0.135M NaOH. 16.93mL of the base were required to reach the endpoint. What is the concentration of the nitric acid solution?

ratio is 1:1 therefore

mol of aicd = mol of base

MV acid = MV base

14.92*M1 = 0.135*16.93

M1 = 0.135*16.93/14.92 = 0.15318699731 M of acid

2)

D = 1.09 g/ml

original solution = 14.92mL

mass = 14.92*1.09 = 16.2628 g of soltion

mol of nitric acid = M*V = 14.92*0.15318699731 = 2.285549999 mmol of nitric acid

mass = mol*MW = 2.285549999*63.01*10^-3 = 144.01250*10^-3= 0.144 g of nitric acid

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