Calculate [H+], [OH -], and the pH for this solution...?

Question

A solution is prepared by mixing 88.0 mL of 5.00 M HCl and 32.0 mL of 8.00 M HNO3. Water is then added until the final volume is 1.00 L. Each answer for [H]+, [OH]-, and pH are worth 1 point, therefore together are worth a total of 3 points

Answer 1

ANSWER:
[H+] = 0.696
[OH-] = 1.44 x 10^-14
pH = 0.157

EXPLAINATION:

88.0ml HCl (1L/1000ml) (5.00molHCl/1L) (1molH+/1molHCl) = 0.44mol H+

32.00ml HNO3 (1L/1000ml) (8.00molHCl/1L) (1molH+/1molHCl) = 0.256 mol H+

Combine the H+ from these two acids:
0.44mol + 0.256mol = 0.696 mol H+

Since the final volume is 1L, the 0.696 mol H+ / 1L = 0.696M H+.

That gives your first answer [H+] = 0.696

To determine [OH-], use eqn:
[H+][OH-] = 1 x 10^-14

[OH-] = (1x 10^-14) / [H+]

[OH-] = (1x 10^-14) / 0.696

[OH-] = 1.44 x 10 -14


Finally to find pH, use eqn:
pH = -log [H+]
pH = -log 0.696
pH = 0.157

Answer 2

88mL * 5M = 0.40 4 mol 7bcfbacd98cfa6ce29d0ad821b14f9Cl 32mL * 8M = 0.256 mol 7bcfbacd98cfa6ce29d0ad821b14f9NO3 using fact the two 7bcfbacd98cfa6ce29d0ad821b14f9Cl and 7bcfbacd98cfa6ce29d0ad821b14f9NO3 are stable acids (aka they dissociate thoroughly into H+ and Cl and NO3), 0.40 4 moles of 7bcfbacd98cfa6ce29d0ad821b14f9Cl at equilibrium might make 0.40 4 moles 7bcfbacd98cfa6ce29d0ad821b14f97bcfbacd98cfa6ce29d0ad821b14f9; and 0.256 moles 7bcfbacd98cfa6ce29d0ad821b14f9NO3 produces 0.256 moles 7bcfbacd98cfa6ce29d0ad821b14f97bcfbacd98cfa6ce29d0ad821b14f9 additionally. so upload those and there are 0.696 moles. divide by using the quantity, 1L, and it relatively is 0.696 M 7bcfbacd98cfa6ce29d0ad821b14f97bcfbacd98cfa6ce29d0ad821b14f9 (first answer). the O7bcfbacd98cfa6ce29d0ad821b14f9- concentration is comparable to the quantity (10^-14)/[7bcfbacd98cfa6ce29d0ad821b14f97bcfbacd98cfa6ce29d0ad821b14f9]. so (10^-14)/(0.696) = a million.40 4 x 10^-14 M for the O7bcfbacd98cfa6ce29d0ad821b14f9- concentration. the p7bcfbacd98cfa6ce29d0ad821b14f9 is the neg-log of 0.696, it relatively is comparable to a p7bcfbacd98cfa6ce29d0ad821b14f9 of 0.157 (it relatively is amazingly low, yet it relatively is with the help of the fact it relatively is stable acids and no bases to counter it)

Answer 3

88mL * 5M = 0.44 mol HCl
32mL * 8M = 0.256 mol HNO3

because both HCl and HNO3 are strong acids (aka they dissociate completely into H and Cl and NO3), 0.44 moles of HCl at equilibrium would make 0.44 moles H+; and 0.256 moles HNO3 produces 0.256 moles H+ also. so add these and there are 0.696 moles. divide by the volume, 1L, and it's 0.696 M H+ (first answer).

the OH- concentration is equal to the quantity (10^-14)/[H+]. so (10^-14)/(0.696) = 1.44 x 10^-14 M for the OH- concentration.

the pH is the neg-log of 0.696, which is equal to a pH of 0.157 (which is very low, but that's because it's strong acids and no bases to counter it)