okay let's do some calculations involving pH and pKa. a zero point zero
five one six molar solution of HNO2 has a pH of two point three four and we want
to know what is the Ka of this acid so we're going to have to do some acid-base
equilibrium calculations this is a relatively weak acid so it's not going
to completely dissociate so we're gonna have to maybe do an icebox here so go
ahead and check out my tutorial on this subject if you need to and when you're
ready give it a shot okay let's write out our acid-base equilibrium we have
nitrous acid reacting with water to yield hydronium and the nitrite ion so
that is the equilibrium we are dealing with and let's go ahead and write out
the Ka expression just like any other equilibrium that is going to involve the
concentrations of the products over the concentrations of the reactants now we
will first note that water is not present in that expression because it is
a pure liquid remember we only put aqueous species and gases in our
equilibrium expressions and we will also note that there are no exponents because
these all have coefficients of one so it is a very simple Ka expression and it
will be important to use so that is what that is
and now as we said we're going to need to set up an icebox
so we've got our initial change in equilibrium and then we have each of the
aqueous species here so let's start out by inputting our initial concentrations
these are the concentrations that were given and so the only one that was given
is zero point zero five one six for nitrous acid so when we say that is the
concentration of the acid that is the moles per liter of acid that exists upon
placing the acid in the solution once it is in solution acid
reactions will begin but at the beginning we have that many moles per
liter of acid and we have no hydronium or nitrite ion so this is prior to any
ionization then immediately upon putting the acid in solution acid-base reactions
begin which means some of the nitrous acid will react
it will deprotonate and for every nitrous acid that deprotonates we get
one hydronium and one nitrite ion so some of the reactant gets used up so the
change for that will be minus X and then the change for the products will be plus
X because for every X that's used up for the reactant we get X for the products
let's also make sure to understand that these come from the stoichiometric
coefficients we have minus 1x and plus 1x and plus 1x because we have a 1 to 1
to 1 ratio for the stoichiometric coefficients from the balanced equation
if those had different coefficients if we had 2 hydronium in the balanced
equation we would have to have plus 2x in the change box for hydronium
but everything is 1 to 1 to 1 so those are going to be the change values then
lastly for the equilibrium values we just take the initial and the change and
add them together so point 0 5 1 6 minus X and then 0 plus X is simply X for the
two products so there is our completed ice box so what can we do with this well
we do have one other piece of information we know that the pH of the
solution is 2 point 3 4 well what is pH pH is just a way of measuring hydronium
concentration specifically the hydronium concentration once equilibrium has been
reached and so we know that the hydronium concentration at equilibrium
is 10 to the negative pH or 10 to the negative 2 point 3 4 and so therefore we
know that the hydronium concentration at equilibrium is zero point zero zero four
six molar that is just expressing the same value that the pH is expressing
just in a different way so we can put zero point zero zero four six in for the
equilibrium value for hydronium that is what we know based on the pH so we can
put that in the table now we know that that was X which means that point zero
zero four six equals x and we can therefore put that in anywhere else we
see X that means the nitrite ion concentration is also point zero zero
four six and then we can get the nitrous acid concentration by taking point zero
five one six minus point zero zero four six which is 0.47 and now it will be
very easy to calculate ka we take our ka expression up there we know it's going
to be hydronium concentration times nitrite concentration over nitrous acid
concentration and that is point zero zero four six times point zero zero four
six divided by point zero four seven those are the numbers we just got from
the equilibrium row of our icebox and we do the arithmetic and the Ka value is
four point five times ten to the negative four.