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

1.19g/ml x 1000ml x 0.38/36.46 = 12M (HCL)

1.42g/ml x 1000ml x 0.7/63.01 = 16M (HN03)
You are OK to here.
(HCL) 50.3 x 12M = 604ml(millimols, not mL) which converts to about 0.6 mol
(HN03) 16.6ml x 16M = 265.6ml (millimols, not mL, which convertw to about 0.266 mol)
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^-).

yaaaaasssssssssssss

To calculate (H+) and (OH-) for the solution, you need to consider the dissociation of the hydrochloric acid (HCl) and nitric acid (HNO3) in water. Both acids are strong acids, meaning they fully dissociate in water.

First, let's calculate the moles of HCl and HNO3 added to the solution:

Moles of HCl = Volume of HCl (in liters) x Concentration of HCl (in moles/liter)
= 0.0503 L x 12 M
= 0.6036 moles HCl

Moles of HNO3 = Volume of HNO3 (in liters) x Concentration of HNO3 (in moles/liter)
= 0.0166 L x 16 M
= 0.2656 moles HNO3

Next, let's calculate the total volume of the solution after adding water:

Total volume of solution = 300 mL + Volume of HCl + Volume of HNO3 + Volume of added water
= 0.300 L + 0.604 L + 0.266 L + Volume of added water
= 1 L

Solving for the Volume of added water:
Volume of added water = Total volume of solution - (Volume of HCl + Volume of HNO3 + 0.300 L)
= 1 L - (0.604 L + 0.266 L + 0.300 L)
= 0.830 L

Now, we can calculate the concentration of (H+) in the solution:

[H+] = Moles of HCl / Total volume of solution
= 0.6036 moles HCl / 1 L
= 0.6036 M

Since HNO3 is a strong acid as well, the concentration of H+ from HNO3 is negligible in comparison to HCl. Therefore, [H+] can be considered 0.6036 M.

The concentration of (OH-) in a solution of water at 25°C is 1 x 10^-7 M (from the autoionization of water).

pH is defined as the negative logarithm (base 10) of the concentration of [H+]. Therefore:

pH = -log10(0.6036) ≈ 0.22 (rounded to two decimal places)

So, the (H+) concentration in the solution is 0.6036 M, the (OH-) concentration is 1 x 10^-7 M, and the pH is approximately 0.22.