FFR: Bases

Bases also have their own special equilibrium constant. It is actually nearly the opposite of the Acid’s constant. For a base, we can define a term called the $pK_b$, which is a function of the equilibrium constant of the base association reaction. For a generic base, Bs, this reaction is given as:

$H_2O+Bs^-\leftrightarrow HBs+OH^-$ $pK_b=-\log K_b$

Notice that this is similar to the Acid dissociation constant, except that in this case, water is used to lend a hydrogen, instead of a lone hydrogen ion being used. Thanks to that, we can determine a very useful relationship between $pK_a$ and $pK_b$ for a conjugate acid-base pair.

$K_a = \frac{[H^+][A^-]}{[HA]}$

$pK_a = -\log\frac{[H^+][A^-]}{[HA]}$

$pK_a=pH -\log\frac{[A^-]}{[HA]}$

$pK_a-pH=-\log\frac{[A^-]}{[HA]}$

$K_b = \frac{[HA][OH^-]}{[A^-]}$

$pK_b=-\log\frac{[HA][OH^-]}{[A^-]}$

$pK_b=pOH-log\frac{[HA]}{[A^-]}$

$pK_b-pOH=-\log\frac{[HA]}{[A^-]}$

$pOH-pK_b=-\log\frac{[A^-]}{[HA]}$

$pK_a-pH = pOH - pK_b$

$pK_a+pK_b=pH+pOH$

$pK_a+pK_b=14$

This is a useful relationship, as it is sometimes preferable to be working with the base equation as opposed to the acid dissociation equation.

Let’s look at what happens when you add a base to water. Our base of choice will be $NH_3$.

$NH_3+H_2O\leftrightarrow NH_4^++OH^-$

We see here that water is providing an extra proton to the Ammonia, converting it into the Ammonium ion, and leaving water one proton less, converting it to the Hydroxide Ion. This reaction details why Ammonia qualifies as an Arrhenius base, even though it doesn’t contain any hydroxide itself.

Like acids, bases can be qualified as either strong or weak. For a base, strong bases are less clearly defined. The idea of a strong base is that it will successfully protonate even a very weak acid. As the Hydroxide ion will always be able to react with an acid, that is a strong base. Furthermore, Hydroxide salts of Alkali and Alkali Earth metals (the first two columns on the periodic table) are all strong bases. They may not be fully soluble, however, whatever does dissolve will dissociate completely, or at least near enough to assume complete dissociation.

Methanol has formula CH₃OH. Its O–H proton and conjugate methoxide CH₃O⁻ should be distinguished from the carbon-bound hydrogen atoms when discussing acid–base behavior.

Example 6.4.1
What is the pH of a 3M solution of NaOH?

That was a strong base though, we didn’t even need an ICE table. Let’s use a weak base this time, and see how the ICE tables work for determining the $pH$ of basic solutions.

Example 6.4.2
What is the pH of a 5M solution of Ammonia ($pK_b=4.75$).

Chapter Problems

1. Order the following solution from lowest $pH$ to highest:
   $1M\ NaOH$ Strong Base
   $0.1M\ HNO_3$ $pK_a=-1.30$
   $0.5M\ NH_3$ $pK_b=4.75$
   $0.5M\ CH_3COOH$ $pK_a=4.76$
2. Reviewed replacement: CO₂(aq)+H₂O(l)⇌H₂CO₃(aq) has hydration equilibrium constant K_h=10⁻²·⁷⁷ in the stated convention. Does K_h determine how quickly hydration occurs?
   Review the authored solution

   No. K_h specifies an equilibrium ratio; kinetic rate constants describe the time dependence. The original label “rate constant pK_h” conflates the two. Full coupled Henry/hydration calculations are retained enrichment.

3. Using the $pK_{a1}$ you calculated in Problem 2, determine the $pH$ of a solution that has 40g of $CO_2$ dissolved into 50L of water. The $pK_{a2}$ of $CO_2=10.33$.
4. Reviewed replacement: a dilute ideal solution of monoprotic weak acid HA has analytical concentration 0.100 M and pH=3.22 at 25 °C. No other acid/base species are present. Neglect water autoionization. Determine pKₐ.
   Review the authored solution

   x=10⁻³·²²=0.00060256 M. Kₐ=x²/(0.100−x)=3.65279×10⁻⁶ and pKₐ=5.43738≈5.44. Supplying analytical concentration closes the original missing-data gap.

5. Three acids are compared in separate beakers. Acid A has a $pK_a=1.3$. Acid B’s conjugate base has a $pK_b=4.44$. A $2M$ solution of Acid C has a $pH=3.3$. Rank the acids from strongest to weakest.
6. What concentration of $H_3PO_4$ is required for the concentration of $[OH^-]$ to become less than $10^{-13}$?
   $pK_{a1}=2.12$
   $pK_{a2}=7.21$
   $pK_{a3}=12.32$
7. Calcium Hydroxide dissolves in water, and as it is a strong base, it fully dissociates:
   $Ca(OH)_2\leftrightarrow Ca^{2+}+2OH^-$. It has a $pK_{sp}=5.26$. Determine the $pH$ of a saturated solution of Calcium Hydroxide.
8. Reviewed replacement: an ideal 25 °C solution has total concentration C=0.00100 M of a stipulated monoprotic weak acid HA with Kₐ=4.30×10⁻⁷. Neglect water autoionization and determine pH. This is an acid-equilibrium model, rather than a prediction for ordinary rainwater.
   Review the authored solution

   Solve Kₐ=x²/(C−x): x=(−Kₐ+√(Kₐ²+4KₐC))/2=2.05226×10⁻⁵ M and pH≈4.688. The original CO₂ pressure and coupled gas/hydration model are not assumed.

9. Sulfuric acid is a diprotic acid, with a strong first dissociation, and a $pK_{a2}=1.99$. If a solution of Sulfuric acid has a $pH = 5.5$, what is the initial concentration of $H_2SO_4$? What mole-percent of the total $H_2SO_4$ exists as $SO_4^{2-}$?
10. Determine the $pOH$ needed to result in $30\%$ deprotonation of $NH_4^+$, which has a $pK_a=9.25$.
11. What mass of Hydrogen Chloride gas is needed to lower the $pH$ of 10L of water to the same as a $1.2M$ solution of $HNO_3$, with a $pK_a=-1.4$.
12. Determine what volume of a $H_3PO_4$ solution with a $pH=3$ is needed to equal the mass of 3L of a $10M$ $NaCl$ solution. Ignore volume increases from addition of solutes.
13. Reviewed replacement: at 25 °C, an ideal solution contains 0.0100 M completely dissociated KOH. Find its pOH and pH.
    Review the authored solution

    [OH⁻]=0.0100 M, so pOH=2.00 and pH=14.00−2.00=12.00. This complete concentration problem replaces the original missing initial amount and unvalidated slow-dissociation model.

14. A solution of multiple acids is at a $pH$ of 2. Given the $pK_a$’s of the acids, order the acids by % de-protonation.
    Acid A: $pK_a=1.11$
    Acid B: $pK_a=5.32$
    Acid C: $pK_a =-2.15$
    Acid D: $pK_a = 7.55$
15. $500mL$ of a $3M$ $HCl$ solution is used to dissolve $20g$ of $Mn(OH)_2$. Determine the $pH$ of the solution after all the Manganese Hydroxide is dissolved.
16. Reviewed replacement: a 20.0 L ideal gas sample is measured at 298.15 K and 1.00 atm. Only its HCl component dissolves into a stipulated final 5.00 L solution, producing pH=2.00 at 25 °C. Treat HCl as fully dissociated; neglect all other reactions. Find HCl mole fraction, using R=0.082057 L atm mol⁻¹ K⁻¹.
    Review the authored solution

    n(HCl)=(10⁻²·⁰⁰)(5.00)=0.050 mol. n(total)=PV/RT=0.817485 mol. The unrounded mole/ideal volume fraction is 0.0611635; report about 6.1% at the two significant figures set by pH 2.00. The revised final volume, temperature and pressure close the original data gaps.

17. What is the concentration of $[SO_4^{2-}]$ in a solution of $0.5M$ $H_2SO_4$?
18. Cyanate, $NCO^-$ is the conjugate base to isocyanic acid, $HNCO$. $5g$ of Potassium Cyanate ($KNCO$) is dissolved into $500mL$ of water. Determine the $pH$ of the solution.
19. Reviewed idealized concentration task at 25 °C: compare the pOH of 4.00 M completely dissociated HCl with the pH of 5.00 M completely dissociated NaOH. Treat concentrations as activities only for this formal exercise.
    Review the authored solution

    HCl: pH=−log₁₀4=−0.602060, pOH=14.602060. NaOH: pOH=−log₁₀5=−0.698970, pH=14.698970. The NaOH pH is higher by 0.096910. Real concentrated solutions require activity treatment.

20. A solution of Phosphoric acid exists as $30\%$ Phosphate. What is the concentration of the Phosphoric acid?

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