Free Printable Acids and Bases Worksheets - Free Printable

Here are the answers to your worksheet, broken down step-by-step.

Part 1: True or False

i. Stronger acids have weaker conjugate bases and weaker acids have stronger conjugate bases.
* Answer: True (T)
* *Reasoning:* Think of it like a tug-of-war. If an acid is "strong," it really wants to give away its proton ($H^+$). Once it gives it away, the leftover piece (the conjugate base) doesn't want it back at all—it is very weak.

ii. Stronger bases have stronger conjugate acids and weaker bases have weaker conjugate acids.
* Answer: False (F)
* *Reasoning:* It works the opposite way! A strong base grabs protons tightly. Its conjugate acid holds onto that proton very tightly and refuses to let go, making it a *weak* acid.

---

Part 2: Write Equations in Water

When writing these equations, remember:
* Acid: Donates an $H^+$ to water ($H_2O$ becomes $H_3O^+$).
* Base: Accepts an $H^+$ from water ($H_2O$ becomes $OH^-$).

i. $F^-$ acts as a base.
$$F^- + H_2O \rightleftharpoons HF + OH^-$$

ii. $HNO_2$ acts as an acid.
$$HNO_2 + H_2O \rightleftharpoons NO_2^- + H_3O^+$$

iii. $Fe(H_2O)_6^{3+}$ acts as an acid.
*(One of the attached water molecules loses an H)*
$$Fe(H_2O)_6^{3+} + H_2O \rightleftharpoons Fe(H_2O)_5(OH)^{2+} + H_3O^+$$

iv. $HCO_3^-$ acts as a base.
$$HCO_3^- + H_2O \rightleftharpoons H_2CO_3 + OH^-$$

v. $HCO_3^-$ acts as an acid.
$$HCO_3^- + H_2O \rightleftharpoons CO_3^{2-} + H_3O^+$$

vi. $Al(H_2O)_5(OH)^{2+}$ acts as a base.
*(It accepts a proton on the OH group to become a water molecule)*
$$Al(H_2O)_5(OH)^{2+} + H_2O \rightleftharpoons Al(H_2O)_6^{3+} + OH^-$$

---

Part 3: Neutralization Calculation

Problem: What volume of $0.45\text{ M LiOH}$ is needed to neutralize $60.0\text{ mL}$ of $0.15\text{ M HI}$?

Step 1: Find moles of Acid (HI).
$$ \text{Moles} = \text{Molarity} \times \text{Volume (in Liters)} $$
$$ 60.0\text{ mL} = 0.060\text{ L} $$
$$ \text{Moles of HI} = 0.15\text{ mol/L} \times 0.060\text{ L} = 0.009\text{ moles} $$

Step 2: Determine ratio.
The reaction is $LiOH + HI \rightarrow LiI + H_2O$. The ratio is 1:1.
So, we need 0.009 moles of LiOH.

Step 3: Calculate Volume of Base (LiOH).
$$ \text{Volume} = \frac{\text{Moles}}{\text{Molarity}} $$
$$ \text{Volume} = \frac{0.009\text{ moles}}{0.45\text{ mol/L}} = 0.02\text{ L} $$

Convert back to mL: $0.02\text{ L} \times 1000 = 20\text{ mL}$.

Answer: 20.0 mL

---

Part 4: Calculate pH Values

i. $0.001\text{ M HCl}$
* HCl is a strong acid, so $[H^+] = 0.001\text{ M}$.
* $pH = -\log(0.001)$
* $0.001 = 10^{-3}$
* pH = 3.0

ii. $0.76\text{ M KOH}$
* KOH is a strong base, so $[OH^-] = 0.76\text{ M}$.
* First find pOH: $pOH = -\log(0.76) \approx 0.12$
* Then find pH: $pH = 14 - pOH$
* $pH = 14 - 0.12$
* pH = 13.88

iii. $2.8 \times 10^{-4}\text{ M Ba(OH)}_2$
* Important: Barium Hydroxide has two OH groups. You must multiply the concentration by 2.
* $[OH^-] = 2 \times (2.8 \times 10^{-4}) = 5.6 \times 10^{-4}\text{ M}$
* Find pOH: $pOH = -\log(5.6 \times 10^{-4}) \approx 3.25$
* Find pH: $pH = 14 - 3.25$
* pH = 10.75

──────────────────────────────────────

Final Answer:

True/False:
i. T
ii. F

Equations:
i. $F^- + H_2O \rightleftharpoons HF + OH^-$
ii. $HNO_2 + H_2O \rightleftharpoons NO_2^- + H_3O^+$
iii. $Fe(H_2O)_6^{3+} + H_2O \rightleftharpoons Fe(H_2O)_5(OH)^{2+} + H_3O^+$
iv. $HCO_3^- + H_2O \rightleftharpoons H_2CO_3 + OH^-$
v. $HCO_3^- + H_2O \rightleftharpoons CO_3^{2-} + H_3O^+$
vi. $Al(H_2O)_5(OH)^{2+} + H_2O \rightleftharpoons Al(H_2O)_6^{3+} + OH^-$

Neutralization Volume:
20.0 mL

pH Values:
i. 3.0
ii. 13.88
iii. 10.75