Karma
May 29
Calculate the volume of 0.100 mol dm-^3 nitric acid needed to react with 20ml of 0.120mol dm-^3 sodium hydroxide than state the pH of these volumes of nitric acid and sodium hydroxide?
and explain why the pH is this value and when more sodium hydroxide is added to the solution what happens to the pH value
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ChatGPT
May 29
To calculate the volume of 0.100 mol dm^-3 nitric acid needed to react with 20ml of 0.120 mol dm^-3 sodium hydroxide, we first need to determine the balanced chemical equation for the reaction between nitric acid (HNO3) and sodium hydroxide (NaOH):
HNO3 + NaOH -> NaNO3 + H2O
From the balanced equation, we can see that the mole ratio between nitric acid and sodium hydroxide is 1:1. Therefore, the number of moles of nitric acid needed to react with 20ml of 0.120 mol dm^-3 sodium hydroxide can be calculated as follows:
Moles of NaOH = concentration x volume
Moles of NaOH = 0.120 mol dm^-3 x 0.020 dm^3
Moles of NaOH = 0.0024 mol
Since the mole ratio between nitric acid and sodium hydroxide is 1:1, we need 0.0024 moles of nitric acid to react with 20ml of 0.120 mol dm^-3 sodium hydroxide.
Now, we can calculate the volume of 0.100 mol dm^-3 nitric acid needed:
Volume of HNO3 = moles / concentration
Volume of HNO3 = 0.0024 mol / 0.100 mol dm^-3
Volume of HNO3 = 0.024 dm^3 or 24 ml
Therefore, 24 ml of 0.100 mol dm^-3 nitric acid is needed to react with 20ml of 0.120 mol dm^-3 sodium hydroxide.
To calculate the pH of the resulting solution after the reaction, we need to consider the neutralization reaction between the acid and base. In this case, the nitric acid and sodium hydroxide will react to form water and sodium nitrate. The resulting solution will be neutral, with a pH of 7.
HNO3 + NaOH -> NaNO3 + H2O
From the balanced equation, we can see that the mole ratio between nitric acid and sodium hydroxide is 1:1. Therefore, the number of moles of nitric acid needed to react with 20ml of 0.120 mol dm^-3 sodium hydroxide can be calculated as follows:
Moles of NaOH = concentration x volume
Moles of NaOH = 0.120 mol dm^-3 x 0.020 dm^3
Moles of NaOH = 0.0024 mol
Since the mole ratio between nitric acid and sodium hydroxide is 1:1, we need 0.0024 moles of nitric acid to react with 20ml of 0.120 mol dm^-3 sodium hydroxide.
Now, we can calculate the volume of 0.100 mol dm^-3 nitric acid needed:
Volume of HNO3 = moles / concentration
Volume of HNO3 = 0.0024 mol / 0.100 mol dm^-3
Volume of HNO3 = 0.024 dm^3 or 24 ml
Therefore, 24 ml of 0.100 mol dm^-3 nitric acid is needed to react with 20ml of 0.120 mol dm^-3 sodium hydroxide.
To calculate the pH of the resulting solution after the reaction, we need to consider the neutralization reaction between the acid and base. In this case, the nitric acid and sodium hydroxide will react to form water and sodium nitrate. The resulting solution will be neutral, with a pH of 7.
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Google Gemini AI
May 29
Step 1: Write the balanced chemical equation.
```
HNO3 + NaOH → NaNO3 + H2O
```
Step 2: Determine the mole ratio from the balanced equation.
```
1 mol HNO3 : 1 mol NaOH
```
Step 3: Calculate the moles of NaOH present.
```
Moles of NaOH = Concentration × Volume
= 0.120 mol dm-³ × 0.020 dm³
= 0.0024 mol
```
Step 4: Calculate the volume of HNO3 required.
```
Moles of HNO3 required = Moles of NaOH
= 0.0024 mol
```
```
Volume of HNO3 required = Moles of HNO3 / Concentration
= 0.0024 mol / 0.100 mol dm-³
= 0.024 dm³
= 24 mL
```
Therefore, 24 mL of 0.100 mol dm-³ nitric acid is required to react with 20 mL of 0.120 mol dm-³ sodium hydroxide.
pH of nitric acid:
```
pH = -log[H+], where [H+] is the molar concentration of hydrogen ions.
```
For a strong acid like nitric acid, the molar concentration of hydrogen ions is equal to the molar concentration of the acid.
```
[H+] = 0.100 mol dm-³
pH = -log(0.100)
= 1.00
```
Therefore, the pH of 24 mL of 0.100 mol dm-³ nitric acid is 1.00.
pH of sodium hydroxide:
For a strong base like sodium hydroxide, the molar concentration of hydroxide ions is equal to the molar concentration of the base.
```
[OH-] = 0.120 mol dm-³
pOH = -log[OH-], where pOH is the negative logarithm of the molar concentration of hydroxide ions.
```
```
pOH = -log(0.120)
= 0.92
```
```
pH + pOH = 14
pH = 14 - pOH
= 14 - 0.92
= 13.08
```
Therefore, the pH of 20 mL of 0.120 mol dm-³ sodium hydroxide is 13.08.
```
HNO3 + NaOH → NaNO3 + H2O
```
Step 2: Determine the mole ratio from the balanced equation.
```
1 mol HNO3 : 1 mol NaOH
```
Step 3: Calculate the moles of NaOH present.
```
Moles of NaOH = Concentration × Volume
= 0.120 mol dm-³ × 0.020 dm³
= 0.0024 mol
```
Step 4: Calculate the volume of HNO3 required.
```
Moles of HNO3 required = Moles of NaOH
= 0.0024 mol
```
```
Volume of HNO3 required = Moles of HNO3 / Concentration
= 0.0024 mol / 0.100 mol dm-³
= 0.024 dm³
= 24 mL
```
Therefore, 24 mL of 0.100 mol dm-³ nitric acid is required to react with 20 mL of 0.120 mol dm-³ sodium hydroxide.
pH of nitric acid:
```
pH = -log[H+], where [H+] is the molar concentration of hydrogen ions.
```
For a strong acid like nitric acid, the molar concentration of hydrogen ions is equal to the molar concentration of the acid.
```
[H+] = 0.100 mol dm-³
pH = -log(0.100)
= 1.00
```
Therefore, the pH of 24 mL of 0.100 mol dm-³ nitric acid is 1.00.
pH of sodium hydroxide:
For a strong base like sodium hydroxide, the molar concentration of hydroxide ions is equal to the molar concentration of the base.
```
[OH-] = 0.120 mol dm-³
pOH = -log[OH-], where pOH is the negative logarithm of the molar concentration of hydroxide ions.
```
```
pOH = -log(0.120)
= 0.92
```
```
pH + pOH = 14
pH = 14 - pOH
= 14 - 0.92
= 13.08
```
Therefore, the pH of 20 mL of 0.120 mol dm-³ sodium hydroxide is 13.08.
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