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pH for any concentration, for strong bases through $K_w$ and pOH, and for weak acids through $K_a$; and why neutral is not always 7.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to calculate pH from a hydronium concentration written in standard form, convert between hydronium, hydroxide, pH and pOH using $K_w$ at 25 °C, find the pH of a strong base and of a weak acid from $K_a$, and explain why pure water at another temperature is neutral at a pH other than 7.
You can read the pH of a concentration that is an exact power of ten, and you know one pH unit is a factor of ten. From the strong and weak acids lesson you have $K_a$ and the ICE table for a weak acid. This lesson handles every other concentration, brings in bases, and shows why pure water is not always at pH 7.
The ion product of water, $K_w$, is the equilibrium constant for water ionising itself: $K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}]$, equal to $1.0 \times 10^{-14}$ at 25 °C. pOH is to hydroxide what pH is to hydronium: $\mathrm{pOH} = -\log_{10}[\mathrm{OH^-}]$. A strong base such as sodium hydroxide gives all its hydroxide ions in solution. A solution is neutral when its hydronium and hydroxide concentrations are equal.
When the concentration is not a power of ten. Write it in standard form, $m \times 10^{-n}$, and use the rule that the logarithm of a product is the sum of the logarithms:
$$\mathrm{pH} = -\log_{10}(m \times 10^{-n}) = n - \log_{10} m$$
For $[\mathrm{H_3O^+}] = 3 \times 10^{-4}$ mol/L, with $\log_{10} 3 = 0.48$, the pH is $4 - 0.48 = 3.52$. The check that catches most mistakes: $3 \times 10^{-4}$ is more than $10^{-4}$, so the solution is a little more acidic than pH 4, and the answer must be a little below 4. A calculator does the logarithm directly; pH is normally given to two decimal places.
When the solution is a base. Water ionises itself very slightly — one molecule gives a proton to another:
$$\mathrm{2H_2O \rightleftharpoons H_3O^+ + OH^-} \qquad K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14}$$
That value is for 25 °C, and the product holds in every water solution, acidic or alkaline. So if you know one ion you know the other, and taking minus the logarithm of both sides gives the handy form, again for 25 °C,
$$\mathrm{pH} + \mathrm{pOH} = 14$$
Sodium hydroxide at $0.01$ mol/L gives $[\mathrm{OH^-}] = 10^{-2}$, so pOH $= 2$ and pH $= 12$.
When the acid is weak. Its hydronium concentration comes from $K_a$. If the acid ionises only slightly, the acid left at equilibrium is almost exactly what was put in, $c$, and the ICE table's expression simplifies:
$$K_a = \frac{x^2}{c - x} \approx \frac{x^2}{c} \quad\Rightarrow\quad [\mathrm{H_3O^+}] = \sqrt{K_a \times c}$$
Ethanoic acid, $K_a = 1.8 \times 10^{-5}$, at $0.10$ mol/L: $\sqrt{1.8 \times 10^{-6}} = 1.3 \times 10^{-3}$ mol/L, pH $2.87$. Hydrochloric acid at the same concentration has pH $1.00$.
Another way: picture
Picture a seesaw with hydronium on one end and hydroxide on the other, balanced so that the product of the two always comes to $10^{-14}$. Push one end up by a factor of a hundred and the other end drops by a factor of a hundred. Neutral is the seesaw level, with both ends at the same height.
Another way: steps
To find the pH of any solution at 25 °C:
$K_w$ is an equilibrium constant, and the ionisation of water is endothermic, so — as the temperature-at-equilibrium lesson predicts — heating water pushes the equilibrium right and raises $K_w$:
| Temperature | $K_w$ | pH of pure water |
|---|---|---|
| 0 °C | $1.1 \times 10^{-15}$ | 7.47 |
| 25 °C | $1.0 \times 10^{-14}$ | 7.00 |
| 50 °C | $5.5 \times 10^{-14}$ | 6.63 |
Pure water at 50 °C has pH 6.63, and it is not acidic. Every hydronium ion in it was made together with one hydroxide ion, so the two concentrations are equal: that is what neutral means. The number 7 is the neutral pH at 25 °C only, which is why the tables in this course say at 25 °C whenever they say 7.
Replacing $c - x$ by $c$ is an approximation, and it is good when the acid is less than about 5% ionised. For ethanoic acid at $0.10$ mol/L, $x = 0.0013$ — about 1.3% — and the exact quadratic gives a pH different only in the third decimal place. For a very dilute acid, or a fairly strong weak acid such as hydrofluoric acid at low concentration, the share ionised grows (dilution pushes the equilibrium towards more ions), the shortcut fails, and the quadratic has to be solved. A quick check after any shortcut: divide $x$ by $c$ and see whether it is small.
A pH answer on the wrong side of the whole number. $5 \times 10^{-3}$ is more hydronium than $10^{-3}$, so its pH is below 3 (it is 2.30), not 3.70.
Using the hydroxide concentration as if it were hydronium. $0.001$ mol/L sodium hydroxide does not have pH 3. Its pOH is 3 and its pH is 11.
Pure water is always pH 7. Only at 25 °C. At other temperatures neutral water has a different pH and is still neutral.
A weak acid's hydronium concentration is its concentration. That treats it as strong. The square root of $K_a \times c$ is far smaller.
pH + pOH = 14 always. It is $\mathrm{p}K_w$, which is 14 only at 25 °C.
Potassium hydroxide solution, $0.020$ mol/L, at 25 °C. Given $\log_{10} 2 = 0.30$.
A strong base: the hydroxide concentration is the base's.
$[\mathrm{OH^-}] = 2 \times 10^{-2}$, so pOH $= 2 - 0.30 = 1.70$.
Standard form, then the logarithm of the product.
pH $= 14 - 1.70 = 12.30$.
Alkaline, and a little above 12 because there is a little more hydroxide than $10^{-2}$.
Methanoic acid, $K_a = 1.6 \times 10^{-4}$ (a rounded value), at $0.10$ mol/L.
A weak acid, so use $K_a$.
$[\mathrm{H_3O^+}] = \sqrt{1.6 \times 10^{-4} \times 0.10} = \sqrt{1.6 \times 10^{-5}} = 4.0 \times 10^{-3}$ mol/L.
$1.6 \times 10^{-5}$ is $16 \times 10^{-6}$, whose root is $4 \times 10^{-3}$.
pH $= 3 - \log_{10} 4 = 3 - 0.60 = 2.40$. Check: $0.004 \div 0.10$ is 4% ionised, so the shortcut holds.
Always check the share ionised after the shortcut.
Hydrochloric acid is strong, so $[\mathrm{H_3O^+}] = 6 \times 10^{-5}$ mol/L.
Every molecule ionised.
pH $= 5 - 0.78 = \ldots$
…$4.22$: a little below 5, as it should be.
lithium hydroxide is a strong base: every unit gives one hydroxide ion. A solution of it at 25 °C has a concentration of $0.01$ mol/L. Put the marker at its pH.
0 |——————————| 14
Mark the position with a cross, then write the value:
A solution has a hydronium ion concentration of $8 \times 10^{-5}$ mol/L. Given that $\log_{10} 8 = 0.90$, what is its pH, to two decimal places?
Answer:
Complete the table for two solutions at 25 °C, where $\mathrm{pH} + \mathrm{pOH} = 14$. Write concentrations as decimals.
| hydronium in mol/L | hydroxide in mol/L | pH | pOH | |
|---|---|---|---|---|
| solution 1 | 0.0001 | |||
| solution 2 | 2 |
In any water solution at 25 °C, $K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 10^{-14}$. Let $h$ be the hydronium ion concentration in mol/L. Write the hydroxide ion concentration as an expression in $h$.
Answer:
A weak acid found in fruit has $K_a = 1 \times 10^{-8}$ at 25 °C. What is the pH of a $0.01$ mol/L solution of it? Assume the share that ionises is too small to change the acid's concentration.
Answer:
At 50 °C the ion product of water is $K_w = 5.5 \times 10^{-14}$, and a sample of pure water at that temperature has a pH of $6.63$. Which statement is correct?
A food factory rinses its equipment with sodium hydroxide solution at pH $11$, at 25 °C. What mass of sodium hydroxide, $\mathrm{NaOH}$, molar mass $40$ g/mol, is dissolved in $3$ L of the rinse?
Answer: unit: g / kg / mg
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A solution at 25 °C has a pH of $9$. What is its hydroxide ion concentration, in mol/L?
Answer: mol/L
You can calculate the pH of any acid or base solution and you know what neutral means. Say out loud why $5 \times 10^{-3}$ mol/L of hydronium gives a pH below 3, not above it. Next: what happens to the pH when an acid and a base meet — neutralisation.
10. Your turn: what is the pH of $6 \times 10^{-5}$ mol/L hydrochloric acid, given $\log_{10} 6 = 0.78$?, step 3