Question

neutralization reactions worksheet
directions: complete the neutralization reaction in the space provided. be sure to balance your
reaction and include states of matter where appropriate. name the acid, base, and salt.
ex
2 hcl (aq)
hydrochloric acid
+
1 ca(oh)₂ (s)
calcium
hydroxide
→
1 cacl₂ (aq)
calcium chloride
+
2 h₂o (l)
water
1
hcn (aq)
+
ba(oh)₂ (s)
→

+
h₂o (l)
water
2
hc₂h₃o₂
(aq)
+
zn(oh)₂ (s)
→

+
h₂o (l)
water
3
hclo (aq)
+
lioh (aq)
→

+
h₂o (l)
water
4
hf (aq)
+
sr(oh)₂ (s)
→

+
h₂o (l)
water
5
h₂s (aq)
+
csoh (aq)
→

+
h₂o (l)
water
6
h₃po₃ (aq)
+
be(oh)₂ (s)
→

+
h₂o (l)
water
7
hno₃ (aq)
+
hg(oh)₂ (s)
→

+
h₂o (l)
water
8
hcl (aq)
+
naoh (aq)
→

+
h₂o (l)
water

Explanation:

Step1: Recall Neutralization Reaction

A neutralization reaction is when an acid (donates \(H^+\)) and a base (donates \(OH^-\)) react to form a salt (ionic compound from acid's anion and base's cation) and water (\(H_2O\)). The general formula is: \( \text{Acid} + \text{Base} ightarrow \text{Salt} + \text{H}_2\text{O} \). We also need to balance the equation and determine states of matter (aq for aqueous, s for solid, l for liquid).

Step2: Solve Reaction 1 (HCN + Ba(OH)₂)

- Acid: HCN (aq, hydrocyanic acid)
- Base: Ba(OH)₂ (s, barium hydroxide)
- Products: Salt (from \(CN^-\) and \(Ba^{2+}\)) and \(H_2O\). The salt is \(Ba(CN)_2\). Now balance:
- HCN has 1 \(H^+\), Ba(OH)₂ has 2 \(OH^-\), so we need 2 HCN to provide 2 \(H^+\) to react with 2 \(OH^-\) (from 1 Ba(OH)₂) to make 2 \(H_2O\).
- Balanced equation: \( 2\text{HCN (aq)} + \text{Ba(OH)}_2 \text{(s)} ightarrow \text{Ba(CN)}_2 \text{(aq)} + 2\text{H}_2\text{O (l)} \)
- Name acid: hydrocyanic acid; base: barium hydroxide; salt: barium cyanide.

Step3: Solve Reaction 2 (HC₂H₃O₂ + Zn(OH)₂)

- Acid: HC₂H₃O₂ (aq, acetic acid)
- Base: Zn(OH)₂ (s, zinc(II) hydroxide)
- Products: Salt (from \(C_2H_3O_2^-\) and \(Zn^{2+}\)) and \(H_2O\). The salt is \(Zn(C_2H_3O_2)_2\). Balance:
- HC₂H₃O₂ has 1 \(H^+\), Zn(OH)₂ has 2 \(OH^-\), so 2 HC₂H₃O₂ to provide 2 \(H^+\) for 2 \(OH^-\) (from 1 Zn(OH)₂) to make 2 \(H_2O\).
- Balanced equation: \( 2\text{HC}_2\text{H}_3\text{O}_2 \text{(aq)} + \text{Zn(OH)}_2 \text{(s)} ightarrow \text{Zn(C}_2\text{H}_3\text{O}_2\text{)}_2 \text{(aq)} + 2\text{H}_2\text{O (l)} \)
- Name acid: acetic acid; base: zinc(II) hydroxide; salt: zinc(II) acetate.

Step4: Solve Reaction 3 (HClO + LiOH)

- Acid: HClO (aq, hypochlorous acid)
- Base: LiOH (aq, lithium hydroxide)
- Products: Salt (from \(ClO^-\) and \(Li^+\)) and \(H_2O\). The salt is LiClO. Balance:
- HClO has 1 \(H^+\), LiOH has 1 \(OH^-\), so 1:1 ratio.
- Balanced equation: \( \text{HClO (aq)} + \text{LiOH (aq)} ightarrow \text{LiClO (aq)} + \text{H}_2\text{O (l)} \)
- Name acid: hypochlorous acid; base: lithium hydroxide; salt: lithium hypochlorite.

Step5: Solve Reaction 4 (HF + Sr(OH)₂)

- Acid: HF (aq, hydrofluoric acid)
- Base: Sr(OH)₂ (s, strontium hydroxide)
- Products: Salt (from \(F^-\) and \(Sr^{2+}\)) and \(H_2O\). The salt is \(SrF_2\). Balance:
- HF has 1 \(H^+\), Sr(OH)₂ has 2 \(OH^-\), so 2 HF to provide 2 \(H^+\) for 2 \(OH^-\) (from 1 Sr(OH)₂) to make 2 \(H_2O\).
- Balanced equation: \( 2\text{HF (aq)} + \text{Sr(OH)}_2 \text{(s)} ightarrow \text{SrF}_2 \text{(aq)} + 2\text{H}_2\text{O (l)} \)
- Name acid: hydrofluoric acid; base: strontium hydroxide; salt: strontium fluoride.

Step6: Solve Reaction 5 (H₂S + CsOH)

- Acid: H₂S (aq, hydrosulfuric acid)
- Base: CsOH (aq, cesium hydroxide)
- Products: Salt (from \(S^{2-}\) and \(Cs^+\)) and \(H_2O\). The salt is \(Cs_2S\). Balance:
- H₂S has 2 \(H^+\), CsOH has 1 \(OH^-\), so 2 CsOH to provide 2 \(OH^-\) to react with 2 \(H^+\) (from 1 H₂S) to make 2 \(H_2O\).
- Balanced equation: \( \text{H}_2\text{S (aq)} + 2\text{CsOH (aq)} ightarrow \text{Cs}_2\text{S (aq)} + 2\text{H}_2\text{O (l)} \)
- Name acid: hydrosulfuric acid; base: cesium hydroxide; salt: cesium sulfide.

Step7: Solve Reaction 6 (H₃PO₃ + Be(OH)₂)

- Acid: H₃PO₃ (aq, phosphorous acid)
- Base: Be(OH)₂ (s, beryllium hydroxide)
- Products: Salt (from \(PO_3^{3-}\) and \(Be^{2+}\)) and \(H_2O\). The salt is \(Be_3(PO_3)_2\). Balance:
- H₃PO₃ has 3 \(H^+\), Be(OH)₂ has 2 \(OH^-\). Find LCM of 3 and 2: 6. So 2 H₃PO₃ (6 \(H^+\)) and 3 Be(OH)₂ (6 \(OH^-\)) to make 6 \(H_2O\).
- Balanced equation: \( 2\text{H}_3\text{PO}_3 \text{(aq)} + 3\text{Be(OH)}_2 \text{(s)} ightarrow \text{Be}_3(\text{PO}_3)_2 \text{(s/aq)} + 6\text{H}_2\text{O (l)} \) (Assume salt is solid or aqueous; here, likely solid as Be(OH)₂ is solid, but check solubility: Be₃(PO₃)₂ is insoluble, so (s).)
- Name acid: phosphorous acid; base: beryllium hydroxide; salt: beryllium phosphite.

Step8: Solve Reaction 7 (HNO₃ + Hg(OH)₂)

- Acid: HNO₃ (aq, nitric acid)
- Base: Hg(OH)₂ (s, mercury(II) hydroxide)
- Products: Salt (from \(NO_3^-\) and \(Hg^{2+}\)) and \(H_2O\). The salt is \(Hg(NO_3)_2\). Balance:
- HNO₃ has 1 \(H^+\), Hg(OH)₂ has 2 \(OH^-\), so 2 HNO₃ to provide 2 \(H^+\) for 2 \(OH^-\) (from 1 Hg(OH)₂) to make 2 \(H_2O\).
- Balanced equation: \( 2\text{HNO}_3 \text{(aq)} + \text{Hg(OH)}_2 \text{(s)} ightarrow \text{Hg(NO}_3\text{)}_2 \text{(aq)} + 2\text{H}_2\text{O (l)} \)
- Name acid: nitric acid; base: mercury(II) hydroxide; salt: mercury(II) nitrate.

Step9: Solve Reaction 8 (HCl + NaOH)

- Acid: HCl (aq, hydrochloric acid)
- Base: NaOH (aq, sodium hydroxide)
- Products: Salt (NaCl) and \(H_2O\). Balance: 1:1 ratio (1 \(H^+\) from HCl, 1 \(OH^-\) from NaOH make 1 \(H_2O\)).
- Balanced equation: \( \text{HCl (aq)} + \text{NaOH (aq)} ightarrow \text{NaCl (aq)} + \text{H}_2\text{O (l)} \)
- Name acid: hydrochloric acid; base: sodium hydroxide; salt: sodium chloride.

Answer (for each reaction, here's the balanced equation, names, and states as an example for Reaction 1):

Reaction 1:

Balanced Equation: \( \boldsymbol{2\text{HCN (aq)} + \text{Ba(OH)}_2 \text{(s)} ightarrow \text{Ba(CN)}_2 \text{(aq)} + 2\text{H}_2\text{O (l)}} \)
Acid: Hydrocyanic acid (HCN)
Base: Barium hydroxide (Ba(OH)₂)
Salt: Barium cyanide (Ba(CN)₂)

(Repeat similar for other reactions following the same steps. For brevity, the key is applying neutralization reaction principles: acid + base → salt + water, balancing by matching \(H^+\) and \(OH^-\) moles, determining salt formula from ions, and naming compounds.)

Answer:

(for each reaction, here's the balanced equation, names, and states as an example for Reaction 1):

Reaction 1:

Balanced Equation: \( \boldsymbol{2\text{HCN (aq)} + \text{Ba(OH)}_2 \text{(s)}
ightarrow \text{Ba(CN)}_2 \text{(aq)} + 2\text{H}_2\text{O (l)}} \)
Acid: Hydrocyanic acid (HCN)
Base: Barium hydroxide (Ba(OH)₂)
Salt: Barium cyanide (Ba(CN)₂)

(Repeat similar for other reactions following the same steps. For brevity, the key is applying neutralization reaction principles: acid + base → salt + water, balancing by matching \(H^+\) and \(OH^-\) moles, determining salt formula from ions, and naming compounds.)