We are given that the concentration of NaOH is 0.0003 M and are asked to calculate the pH
We know that NaOH dissociates by the following reaction:
NaOH → Na⁺ + OH⁻
Which means that one mole of NaOH produces one mole of OH⁻ ion, which is what we care about since the pH is affected only by the concentration of H⁺ and OH⁻ ions
Now that we know that one mole of NaOH produces one mole of OH⁻, 0.0003M NaOH will produce 0.0003M OH⁻
Concentration of OH⁻ (also written as [OH⁻]) = 3 * 10⁻⁴
<u>pOH of the solution:</u>
pOH = -log[OH⁻] = -log(3 * 10⁻⁴)
pOH = -0.477 + 4
pOH = 3.523
<u>pH of the solution:</u>
We know that the sum of pH and pOH of a solution is 14
pH + pOH = 14
pH + 3.523 = 14 [subtracting 3.523 from both sides]
pH = 10.477
Let the 8% solution be A, the 20% solution be B and the final solution be C.
C = A + B
C = 12 + B
0.16C = 0.08(12) + 0.2(B)
0.16(12 + B) = 0.96 + 0.2B
0.96 = 0.04B
B = 24 Liters
C = 12 + 24
C = 36 Liters
There are 21 atoms represented in the formula :) hope this helped.
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Answer:
% Mg = 16.2 % ≈ 16 %
%N = 18.9 % ≈ 19 %
% O = 64.9 % ≈ 65 %
Explanation:
Step 1: Data given
Molar mass Mg(NO3)2 = 148 g/mol
Molar mass Mg = 24 g/mol
Molar mass N = 14 g/mol
Molar mass O = 16 g/mol
Step 2: Calculate the percent composition
% Mg = (24/148) *100%
% Mg = 16.2 % ≈ 16 %
% N = (2*14 / 148) * 100%
%N = 18.9 % ≈ 19 %
% O = (6*16/148) * 100%
% O = 64.9 % ≈ 65 %
