19. 158 mL of a 0.148M NaCI solution is added to 228 mL of a 0.369M NHANOs solut

Question

19. 158 mL of a 0.148M NaCI solution is added to 228 mL of a 0.369M NHANOs solution. The concentration of ammonium ions in the resulting mixture is 20. 1.40 g of silver nitrate is dissolved in 125 mL of water. To this solution is added 5.00 mL of 1.50M hydrochloric acid, and a precipitate forms. Find the concentration of silver ions remaining in solution.. x1o3M 21. Calcium sulfate dihydrate (commonly known as gypsum) dissolves in cold water to the extent of 0.241 g per 100. cm3. What is the concentration of calcium ions in this solution? 22. Calcium nitrate tetrahydrate dissolves in cold water to the extent of 266 g per 100. cm3 What is the concentration of nitrate ions in this solution?

Explanation / Answer

19.By adding a solution of NaCl tO the solution of NH4NO3 you just change the volume and concentration of NH4NO3 because NaCl and NH4NO3 donot react with each other.

1mol of NH4NO3 gives in dissociation process 1mol of NH4+.
To find the new concentration you need to find the number of moles of NH4NO3

Moles(NH4NO3)=Mx V=0.369mol/L x 0.228L= 0.084 mole

New concentration of NH4NO3 or NH4+ is calculated by dividing the number of moles with the sum of volumes of two solutions
M(NH4+)=n/(V(NH4NO3) + V(NaCl))

=0.084/(0.158+0.228)

=0.084/0.386

=0.218 M

22. moles Ca(NO3)2 * 4 H2O = 266 g / 236.15 g/mol=1.13

Moles NO3- = 2 x 1.13 = 2.26

concentration NO3- = 2.26 / 0.100 L = 22.6 M

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