(1) Let’s write out the chemical reaction to see what we are looking at.
NaOH is one of the strong bases we need to have memorized for chemistry.
We know that strong bases FULLY DISSOCIATE into their ions when they react with water, so the product should include all ions and some leftover H20 and heat.
so;
NaOH(s) + H2O(l) => Na+ + OH- + H20 + HEAT
We are asked to find the pH, which is equal to the negative log of the concentration of H+ ions. Our reaction only shows us the product of OH- ions.
We know;
pH = -log (H+)
and
pOH = -log (OH-)
And we know that we can describe any substance using the equation;
pH + pOH = 14
So, we can find the pOH and use it to find our pH.
pOH = -log (OH-)
we have a strong base w/ a concentration of 1.5 M (mols/liter), so the concentration of ions in the reaction will be the same as the beginning solution.
pOH = -log (1.5)
pOH= -0.176091259
pH + pOH = 14
pH + -0.176091259 = 14
pH = 14.17609 =~ 14.18
(2) Here we are given the pOH, which is the negative log of the concentration of OH- ions, and we are asked to find the concentration of H+ ions, which is equal to 10pH.
H3O+ refers to the concentration of H+ ions, they’re just bound to water molecules.
we are given that our pOH = 10.6
pH + pOH = 14
10.6 + pH = 14
pH = 3.4
pH = 3.4, and pH = -log (H+)
3.4 = -log (H+)
log-1(3.4) = H+
concentration of H+ = 3.98 x 10-4 M (mols/liter)
(3) Benzoic acid = C6H5CO2H
This is NOT one of our strong acids, so it will not completely dissociate in water. There will be some ions, but there will still be benzoic acid present in the final product.
We can estimate how much a weak acid will dissociate by its dissociation constant, Ka, which is unique to each weak acid.
We are given that benzoic acid has a Ka = 6.5 x 10-5
The formula for Ka is:
Ka = (H+)(B-) / (HB)
H+ = concentration of hydrogen ions, B- = concentration of conjugate base ions, HB = concentration of unchanged acid molecules
Our acid is formed by the ions;
C6H5COO- + H+
So we know that ONE conjugate base ion forms for every ONE H+ ion present
Since our HB = concentration of unchanged molecules, we can also say that it is equal to the initial concentration – the final concentration of H+ molecules. We gain an H+ for every initial acid molecule we break down.
So;
let x= concentration of H+ in product
Then our HB = initial concentration benzoic acid – x;
HB = 1.5 M – x
Now, we will replace our HB term in our Ka equation with the with our new values;
[Ka] = 6.5 x 10-5 ,[ HB ]= 1.5 – x , [H+ & B-] = xKa = (H+)(B-) / (HB)
6.5 x 10-5 = (x * x) / (1.5 – x)
now we move everything to the same side of the equation and write in quadratic form;
x2 + 6.5 x 10-5x + 9.75 x 10-5 = 0
When we solve using the quadratic formula, we find that x is so small it is considered negligible to the Ka equation.
So, we can use the equation that;
Ka = x2 / 1.5 M
because the subtraction of “x” ions in the denominator is negligible to the equation.
so,
6.5 x 10-5 = x2 / 1.5
x2 = 9.75 x 10-5
x = 9.874 x 10-3 H+ ions
pH = -log ( H+)
pH = -log (9.874 x 10-3) = 2.0055068
pH = ~ 2.0