A solution is prepared by adding 50 mL of .050 M HCL to 150 mL of .10 M HNO3. Calculate the concentrations. I know I need to find the [H+] and [OH-] but im not sure how.
moles HNO3 = M x L = ??
M HNO3 = moles/total L.
moles HCl = M x L = xx
M HCl = moles/total L.
Another way, to me less complicated; however, it gets away from the defnitions of M = moles/L which is all the above. Just the definition. The other way is a dilution method.
(HNO3) = 0.1M x (150 mL/200 mL) = ?M
(HCl) = 0.05M x (50 mL/200 mL) = ?M
Try it both ways. You should get the same answer either way.
To find the concentration of H+ in the solution, we can start by calculating the number of moles of HCl and HNO3 present in the solution.
First, let's find the moles of HCl:
Moles of HCl = volume (L) × concentration (mol/L)
Moles of HCl = 0.050 L × 0.050 mol/L
Moles of HCl = 0.0025 mol
Next, let's find the moles of HNO3:
Moles of HNO3 = volume (L) × concentration (mol/L)
Moles of HNO3 = 0.150 L × 0.10 mol/L
Moles of HNO3 = 0.015 mol
Now, we can add the moles of HCl and HNO3 to get the total moles of H+ in the solution:
Total moles of H+ = moles of HCl + moles of HNO3
Total moles of H+ = 0.0025 mol + 0.015 mol
Total moles of H+ = 0.0175 mol
Finally, we need to calculate the concentration of H+ in the solution:
Concentration (mol/L) = moles of H+ / volume (L)
Concentration of H+ = 0.0175 mol / 0.200 L
Concentration of H+ = 0.0875 mol/L or 0.0875 M
To calculate the concentration of OH-, we need to use the fact that water undergoes autoionization to produce H+ and OH-. At 25°C, the autoionization constant for water (Kw) is 1.0 × 10^-14.
Since the solution is acidic, the concentration of OH- will be less than the concentration of H+. It can be approximated by assuming that the concentration of OH- is equal to the concentration of HCl (which is much smaller than the concentration of HNO3).
Therefore, the concentration of OH- can be approximated as 0.050 mol/L or 0.050 M.
To calculate the concentrations of [H+] and [OH-] in the resulting solution, you need to consider the ionization reactions of both acids: HCl and HNO3.
1. For HCl:
HCl (aq) → H+ (aq) + Cl- (aq)
Since HCl is a strong acid, it fully dissociates in water, meaning that it completely ionizes into H+ and Cl-. Therefore, the concentration of H+ is equal to the initial concentration of HCl, which is 0.050 M.
2. For HNO3:
HNO3 (aq) → H+ (aq) + NO3- (aq)
Similar to HCl, HNO3 is also a strong acid and completely dissociates in water. Thus, the concentration of H+ is equal to the initial concentration of HNO3, which is 0.10 M.
Since both HCl and HNO3 provide H+ ions, you can simply add the two concentrations to obtain the total concentration of H+ in the solution.
[H+] total = [H+] HCl + [H+] HNO3 = 0.050 M + 0.10 M = 0.15 M
Since water is neutral, the concentration of [OH-] in the solution can be calculated using the relationship:
[H+][OH-] = 1.0 x 10^-14 at 25°C
Here, [H+] is 0.15 M (as calculated above). Rearranging the equation, you can solve for [OH-]:
[OH-] = 1.0 x 10^-14 ÷ [H+]
= 1.0 x 10^-14 ÷ 0.15
= 6.67 x 10^-14 M
Therefore, the concentrations in the resulting solution are [H+] = 0.15 M and [OH-] = 6.67 x 10^-14 M.
The balanced chemical equation for reaction of Ba(OH) and HCl is as follows:
Ba(OH) + 2HCl BaCl + 2H O
1 mole Ba(OH) produce 2 moles OH ions.
Number of moles of OH ion in solution = concentration of Ba(OH) x volume of Ba(OH)
= 0.1 mol L x 0.03 L x 2
= 0.006 mol
Number of moles of H ion in solution = Concentration of HCl x volume of HCl
= 0.05 mol L x 0.02 L
= 0.001 mol
Now after mixing 0.001 mol H will reacts with 0.001 mol OH .
Number of moles of OH left unreacted = 0.006 - 0.001
= 0.005 mol
Total volume of solution = 20 ml + 30 ml = 50 ml = 0.05 L
Concentration of OH ion in ánal solution = 0.005 mol / 0.05 L = 0.1 mol L .
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