Answer:
Molarity of the sodium hydroxide solution is 1.443 M/L
Explanation:
Given;
0.60 M concentration of NaOH contains 2.0 L
3.0 M concentration of NaOH contains 495 mL
Molarity is given as concentration of the solute per liters of the solvent.
If the volumes of the two solutions are additive, then;
the total volume of NaOH = 2 L + 0.495 L = 2.495 L
the total concentration of NaOH = 0.6 M + 3.0 M = 3.6 M
Molarity of NaOH solution = 3.6 / 2.495
Molarity of NaOH solution = 1.443 M/L
Therefore, molarity of the sodium hydroxide solution is 1.443 M/L
The concentration may be expressed as % m/m, this is the mass of ions in 100 mass units of solution, whose formula is:
% m/m = [mass of ions / mass of solution]*100
Then,
%m/m = [8.5*10^ -3 grams of calcium ions] / [490 grams of solution] * 100 =
% m/m = 1.74 * 10^ -5 %
Answer: 1.74 * 10 ^ -5 %
Answer:
Kₐ = 6.7 x 10⁻⁴
Explanation:
First lets write the equilibrium expression, Ka , for the dissociation of hydrofluoric acid:
HF + H₂O ⇄ H₃O⁺ + F⁻
Kₐ = [ H₃O⁺ ] [ F⁻ ] /[ [ HF ]
Since we are given the pH we can calculate the [ H₃O⁺ ] ( pH = - log [ H₃O⁺ ] , and because the acid dissociates into a 1: 1 relation , we will also have [F⁻ ]. The [ HF ] is given in the question so we have all the information that is needed to compute Kₐ.
pH = -log [ H₃O⁺ ]
1.68 = - log [ H₃O⁺ ]
Taking antilog to both sides of this equation:
10^-1.68 = [ H₃O⁺ ] ⇒ 2.1 X 10⁻² M= [ H₃O⁺ ]
[ F⁻ ] = 2.1 X 10⁻² M
Solving for Kₐ :
Kₐ = ( 2.1 X 10⁻² ) x ( 2.1 X 10⁻² ) / 0.65 = 6.7 x 10⁻⁴
(Rounded to two significant figures, the powers of 10 have infinite precision )
Answer:
Pure water is a non conductor of electricity and dilute acids in their aqueous solutions form free ions, which conducts electricity. Thus when we need to electrolyse water, a dilute acid is added to increase its conductivity.
