Find the molarity of [H+] and [OH-] in the following solutions:
a) 0.050 molar NaOH
b) 2.09g of calcium hydroxide in 100 mL of water
c) 0.875 M H2SO4
a)(OH^-) = 0.05.
(H^+)(OH^-) = Kw = 1E-14. Substitute and solve for (H^+)
b) Convert grams Ca(OH)2 to mols, convert that to mols (OH^-), then follow part a.
Thanks!
To find the molarity of [H+] and [OH-] in a solution, we need to determine the concentration of hydrogen ions ([H+]) and hydroxide ions ([OH-]).
a) For 0.050 molar NaOH:
NaOH dissociates completely in water to form Na+ and OH- ions. Thus, the concentration of [OH-] in the solution is equal to the molarity of NaOH. Therefore, [OH-] = 0.050 M.
[H+] can be calculated using the fact that water is a neutral substance. Since [H+] = [OH-] in a neutral solution, [H+] is also 0.050 M.
b) For 2.09g of calcium hydroxide in 100 mL of water:
First, we need to convert the grams of calcium hydroxide into moles. The molar mass of calcium hydroxide (Ca(OH)2) is 74.1 g/mol.
Moles of Ca(OH)2 = Mass (g) / Molar mass (g/mol) = 2.09g / 74.1 g/mol = 0.0282 mol.
Now, we need to determine the number of moles in the 100 mL solution:
Molarity (M) = Moles / Volume (L)
Volume = 100 mL = 100/1000 L = 0.1 L
Molarity of [OH-] = Moles / Volume = 0.0282 mol / 0.1 L = 0.282 M.
[H+] can be found using the fact that water is neutral, so [H+] = [OH-]. Therefore, [H+] = 0.282 M.
c) For 0.875 M H2SO4:
H2SO4 is a strong acid that fully dissociates in water. It releases 2 moles of H+ ions for every mole of H2SO4.
Therefore, [H+] = 2 * Molarity of H2SO4 = 2 * 0.875 M = 1.75 M.
Since water is neutral, [OH-] = 1.75 M.
By following these calculations, we can find the molarity of [H+] and [OH-] in the given solutions.