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To solve the titration problems, we need to use the concept of stoichiometry and the relationship between molarity, volume, and the neutralization reaction. The main formula we use is:
[tex]\[ M_1 V_1 = M_2 V_2 \][/tex]
where:
- [tex]\( M_1 \)[/tex] and [tex]\( V_1 \)[/tex] are the molarity and volume of the acid.
- [tex]\( M_2 \)[/tex] and [tex]\( V_2 \)[/tex] are the molarity and volume of the base.
### Part (a): 25 mL of 2.43 M HCl solution
For hydrochloric acid (HCl) and sodium hydroxide (NaOH) titration, the reaction is as follows:
[tex]\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \][/tex]
This reaction is a 1:1 reaction, meaning one mole of HCl reacts with one mole of NaOH.
Given:
- Volume of HCl ([tex]\( V_{\text{HCl}} \)[/tex]) = 25 mL
- Molarity of HCl ([tex]\( M_{\text{HCl}} \)[/tex]) = 2.43 M
- Molarity of NaOH ([tex]\( M_{\text{NaOH}} \)[/tex]) = 1.42 M
Using the formula [tex]\( M_1 V_1 = M_2 V_2 \)[/tex]:
[tex]\[ (2.43 \, \text{M}) \times (25 \, \text{mL}) = (1.42 \, \text{M}) \times V_{\text{NaOH}} \][/tex]
Solving for [tex]\( V_{\text{NaOH}} \)[/tex]:
[tex]\[ V_{\text{NaOH}} = \frac{(2.43 \, \text{M}) \times (25 \, \text{mL})}{1.42 \, \text{M}} = 42.78 \, \text{mL} \][/tex]
### Part (b): 25 mL of 1.5 M H3PO4 solution
For phosphoric acid (H3PO4) and sodium hydroxide (NaOH) titration, the reaction is as follows:
[tex]\[ \text{H}_3\text{PO}_4 + 3\text{NaOH} \rightarrow \text{Na}_3\text{PO}_4 + 3\text{H}_2\text{O} \][/tex]
This reaction is a 1:3 reaction, meaning one mole of H3PO4 reacts with three moles of NaOH.
Given:
- Volume of H3PO4 ([tex]\( V_{\text{H3PO4}} \)[/tex]) = 25 mL
- Molarity of H3PO4 ([tex]\( M_{\text{H3PO4}} \)[/tex]) = 1.5 M
- Molarity of NaOH ([tex]\( M_{\text{NaOH}} \)[/tex]) = 1.42 M
Since H3PO4 has 3 protons, we need to multiply the molarity by 3:
[tex]\[ M_{\text{H3PO4, equivalent}} = 1.5 \, \text{M} \times 3 = 4.5 \, \text{M} \][/tex]
Using the formula [tex]\( M_1 V_1 = M_2 V_2 \)[/tex]:
[tex]\[ (4.5 \, \text{M}) \times (25 \, \text{mL}) = (1.42 \, \text{M}) \times V_{\text{NaOH}} \][/tex]
Solving for [tex]\( V_{\text{NaOH}} \)[/tex]:
[tex]\[ V_{\text{NaOH}} = \frac{(4.5 \, \text{M}) \times (25 \, \text{mL})}{1.42 \, \text{M}} = 79.23 \, \text{mL} \][/tex]
Therefore, the volumes of 1.42 M NaOH solution required to titrate the given solutions are:
(a) 42.78 mL for 25 mL of 2.43 M HCl solution
(b) 79.23 mL for 25 mL of 1.5 M H3PO4 solution
[tex]\[ M_1 V_1 = M_2 V_2 \][/tex]
where:
- [tex]\( M_1 \)[/tex] and [tex]\( V_1 \)[/tex] are the molarity and volume of the acid.
- [tex]\( M_2 \)[/tex] and [tex]\( V_2 \)[/tex] are the molarity and volume of the base.
### Part (a): 25 mL of 2.43 M HCl solution
For hydrochloric acid (HCl) and sodium hydroxide (NaOH) titration, the reaction is as follows:
[tex]\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \][/tex]
This reaction is a 1:1 reaction, meaning one mole of HCl reacts with one mole of NaOH.
Given:
- Volume of HCl ([tex]\( V_{\text{HCl}} \)[/tex]) = 25 mL
- Molarity of HCl ([tex]\( M_{\text{HCl}} \)[/tex]) = 2.43 M
- Molarity of NaOH ([tex]\( M_{\text{NaOH}} \)[/tex]) = 1.42 M
Using the formula [tex]\( M_1 V_1 = M_2 V_2 \)[/tex]:
[tex]\[ (2.43 \, \text{M}) \times (25 \, \text{mL}) = (1.42 \, \text{M}) \times V_{\text{NaOH}} \][/tex]
Solving for [tex]\( V_{\text{NaOH}} \)[/tex]:
[tex]\[ V_{\text{NaOH}} = \frac{(2.43 \, \text{M}) \times (25 \, \text{mL})}{1.42 \, \text{M}} = 42.78 \, \text{mL} \][/tex]
### Part (b): 25 mL of 1.5 M H3PO4 solution
For phosphoric acid (H3PO4) and sodium hydroxide (NaOH) titration, the reaction is as follows:
[tex]\[ \text{H}_3\text{PO}_4 + 3\text{NaOH} \rightarrow \text{Na}_3\text{PO}_4 + 3\text{H}_2\text{O} \][/tex]
This reaction is a 1:3 reaction, meaning one mole of H3PO4 reacts with three moles of NaOH.
Given:
- Volume of H3PO4 ([tex]\( V_{\text{H3PO4}} \)[/tex]) = 25 mL
- Molarity of H3PO4 ([tex]\( M_{\text{H3PO4}} \)[/tex]) = 1.5 M
- Molarity of NaOH ([tex]\( M_{\text{NaOH}} \)[/tex]) = 1.42 M
Since H3PO4 has 3 protons, we need to multiply the molarity by 3:
[tex]\[ M_{\text{H3PO4, equivalent}} = 1.5 \, \text{M} \times 3 = 4.5 \, \text{M} \][/tex]
Using the formula [tex]\( M_1 V_1 = M_2 V_2 \)[/tex]:
[tex]\[ (4.5 \, \text{M}) \times (25 \, \text{mL}) = (1.42 \, \text{M}) \times V_{\text{NaOH}} \][/tex]
Solving for [tex]\( V_{\text{NaOH}} \)[/tex]:
[tex]\[ V_{\text{NaOH}} = \frac{(4.5 \, \text{M}) \times (25 \, \text{mL})}{1.42 \, \text{M}} = 79.23 \, \text{mL} \][/tex]
Therefore, the volumes of 1.42 M NaOH solution required to titrate the given solutions are:
(a) 42.78 mL for 25 mL of 2.43 M HCl solution
(b) 79.23 mL for 25 mL of 1.5 M H3PO4 solution
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