Question

A solution of is prepared by adding 50.3ml of concentrated hydrochloric acid and 16.6ml of concentrated nitric acid to 300ml of water. More water is added until the final volume is 1.00L.
Calculate (H+) (OH) and the PH for this solution. Hint concentrated HCL is 38% hcl by mass and has a density 1.19g/ml.
Concentrated HN03 is 70% HN03 and has a density of 1.42g/ml

My calculation
1.19g/ml x 1000ml x 0.38/36.46 = 12M (HCL)
1.42g/ml x 1000ml x 0.7/63.01 = 16M (HN03)
(HCL) 50.3 x 12M = 604ml
(HN03) 16.6ml x 16M = 265.6ml

I am not completed. Who helps me to solve it for me. Pls help

Answer 1

1.19g/ml x 1000ml x 0.38/36.46 = 12M (HCL)
1.42g/ml x 1000ml x 0.7/63.01 = 16M (HN03)
<b>You are OK to here.</b>
(HCL) 50.3 x 12M = 604ml<b>(millimols, not mL) which converts to about 0.6 mol</b>
(HN03) 16.6ml x 16M = 265.6ml <b>(millimols, not mL, which convertw to about 0.266 mol)</b>
Then M = total mols/total L or about
0.87 mol/1.00L = about 0.87M

Since both HCl and HNO3 are strong acids they ionize 1oo%; therefore, (H^+) = same as molarity of the solution. pH is obtained from this. Then (H^+)(OH^-) = Kw = 1E-14. You know (H^+) and Kw, solve for (OH^-).

Answer 2

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