conc HCL=2.0E-5/2=1E-5
pH=-log concH+=5
now, diltued 3x as above
pH=-log1/3 E-5=- log .333E-5=5.48
Consider a 500mL solution containing 2.0×10^-5 mol HCl
·what is the pH of the above solution?
·If the solution is diluted to 1.5 L what will be the new pH?
3 answers
2 x 10⁻⁵mole HCl in 500-ml Solution:
=>[HCl] = [H₃O⁺] = [H⁺] = (moles H⁺ / volume in Liters)
= (2 x 10⁻⁵ mole H⁺/ 0.500 Liters) = 4.0 x 10⁻⁵ Molar in H⁺
pH = -log[H₃O⁺] = -log[H⁺] = -log(4 x 10⁻⁵) = 4.4
Diluting to 1.5 Liters:
=> [HCl] = [H₃O⁺] = [H⁺] = (moles H⁺ / volume in Liter) = (2 x 10⁻⁵ mole H⁺/ 1.500 Liters) = 1.33 x 10⁻⁵M in H⁺
pH = -log[H₃O⁺] = -log[H⁺] = -log(1.33 x 10⁻⁵) = 4.9
=>[HCl] = [H₃O⁺] = [H⁺] = (moles H⁺ / volume in Liters)
= (2 x 10⁻⁵ mole H⁺/ 0.500 Liters) = 4.0 x 10⁻⁵ Molar in H⁺
pH = -log[H₃O⁺] = -log[H⁺] = -log(4 x 10⁻⁵) = 4.4
Diluting to 1.5 Liters:
=> [HCl] = [H₃O⁺] = [H⁺] = (moles H⁺ / volume in Liter) = (2 x 10⁻⁵ mole H⁺/ 1.500 Liters) = 1.33 x 10⁻⁵M in H⁺
pH = -log[H₃O⁺] = -log[H⁺] = -log(1.33 x 10⁻⁵) = 4.9
Dr Rebel is correct, I divided 2 by .5 and got 1. UGH