A solution is prepared by adding 50 ml of 0.050 M HCl to 150mlof 0.10 M HNO 3 .

Question

A solution is prepared by adding 50 ml of 0.050 M HCl to 150mlof 0.10 M HNO3. Calculate the concentrationsof all species in this solution. I know the speciespresent, but I can not figure out how the book comes up with eachsolution.
Answers:                                              [H+]  = 0.088 M [OH-] = 1.1*10-13 M [Cl-] = 0.013 M [NO3-] = 0.075 M A solution is prepared by adding 50 ml of 0.050 M HCl to 150mlof 0.10 M HNO3. Calculate the concentrationsof all species in this solution. I know the speciespresent, but I can not figure out how the book comes up with eachsolution.
Answers:                                              [H+]  = 0.088 M [OH-] = 1.1*10-13 M [Cl-] = 0.013 M [NO3-] = 0.075 M

Explanation / Answer

Total volume = 50 + 150 = 200mL = 0.2L n(HCl) = cv = 0.05 x 0.05 = 0.0025mol n(HNO3) = cv = 0.1 x 0.15 = 0.015mol 1 mol of HCl contains 1 mol of H+ 1 mol of HNO3 contains 1 mol of H+ Total n(H+) = 0.0025 + 0.015 = 0.0175mol [H+] = 0.0175/0.2 = 0.088M [H+][OH-] = 1 x 10-14 [OH-] = ( 1 x 10-14) / (0.088) = 1.1 x10-13M 1 mol of HCl contains 1 mol of Cl- n(Cl-) = 0.0025mol [Cl-] = 0.0025/0.2 = 0.013M 1 mol of HNO3 contains 1 mol ofNO3- n(NO3-) = 0.015mol [NO3-] = 0.015/0.2 = 0.075M Hope this helps!

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