A solution is prepared by adding 42.0 mL of 0.060 M HCl to 154.0 mL of 0.10 M HNO3.?

Question

calculate the concentrations of all species in this solution.
[H+] mol/liter
[OH -] mol/liter
[Cl -] mol/liter
[NO-] mol/liter

i found that:
H+ =.091
CL=.013
NO3=.079

but none of my OH answers have worked yet, please help/explain
thanks!

Answer 1

The total volume of the solution is 42+154 =196mL =0.196L
the dissodiation= of HCl is HCl--->H+ + Cl-
and HNO3-->H+ +NO3-
The molarity of H+
millimole of H+ : 60*42+0.1*154=2.52+15.4=17.6/0.196=89.8mM/L =0.898 mole/liter
millimole Cl- =15.4/0.196=0.079mole/L

Answer 2

New [HCl] = [42.0/(42.0+154.0)] * (0.060) = 0.01286mol/mL
New [HNO3] = [152.0/(42.0+154.0)] * (0.10) = 0.07857mol/mL

In pure water, [H+] = [OH-] = 10^-7 mol/mL
Well, if you want to count them in... But it is usually ignored...

[H+] = 0.01286 + 0.07857 + 10^-7 = (0.09143 + 10^-7) mol/mL
[OH-] = 10^-7 mol/mL
[Cl-] = 0.01286mol/mL
[NO3-] = 0.07857mol/mL

Answer 3

"i found that:
H+ =.091"

If this is true then pH = -log(.091) = 1.041
pH + pOH = 14 so pOH = 14 - 1.041 = 12.959
pOH = -log(OH) so OH = 10^(-pH) = 10^(-12.959) = 1.1e-13
so OH- = 1.1e-13