Remember pH is defined as: pH = -log[H+]
so, substituting
4 = - log[H+]
-4 = log [H+] Take antilog of both sides
10^(-4) = [H+]
[H+] = 1 x 10^-4 M
Answer:
Molarity of the solution = 3.000 M
Volume of the solution = 250.0 mL = 0.25 L
moles in 250.0 mL = molarity x volume of the solution
= 3.000 M x 0.25 L
= 0.75 mol
Hence, 0.75 mol of NaCl is needed to prepare 250.0 mL of 3.000 M NaCl solution.
Moles (mol) = mass (g) / molar mass (g/mol)
Moles of NaCl in 250.0 mL = 0.75 mol
Molar mass of NaCl = 58.44 g/mol
Mass of NaCl in 250.0 mL = Moles x Molar mass
= 0.75 mol x 58.44 g/mol
= 43.83 g
Hence, 43.83 g of NaCl is needed to prepare 250.0 mL of 3.000 M solution.
Explanation:
The liquid that is use to dissolve solute
