11.During an acid-base titration, 25 mL of NaOH 0.2 M were required to neutralize 20 mL of HCl.
Calculate the pH of the solution for each of the following:
12.Before the titration.
13.After adding 24.9 mL of NaOH.
14.At the equivalence point.
15.After adding 25.1 mL of NaOH.
16.The pH of the equivalence point during a certain titration is 3.5.
17.Which indicator would be adequate for this titration?
18.Which indicator would you suggest to be avoided during this titration? Justify.
19.Calculate the volume of NaOH 0.5 mol/L necessary to neutralize 300 mL of HCl 0.2 M.
12. Before the titration, the pH of the solution can be calculated using the concentration of the HCl solution. However, since the concentration of the HCl solution is not provided in the question, it is not possible to calculate the exact pH.
13. After adding 24.9 mL of NaOH, we can calculate the number of moles of NaOH used using the molarity and volume of NaOH added.
Moles of NaOH = Molarity * Volume
= 0.2 mol/L * 0.0249 L
= 0.00498 moles
Since NaOH reacts in a 1:1 ratio with HCl, the number of moles of HCl neutralized in the reaction is also 0.00498 moles.
To calculate the concentration of HCl remaining, we can use the volume of HCl initially and the moles of HCl neutralized.
Molarity of HCl remaining = Moles of HCl remaining / Volume of HCl remaining
= (Moles of HCl initially - Moles of NaOH) / (Volume of HCl initially - Volume of NaOH added)
= (0.2 mol/L * 0.02 L - 0.00498 moles) / (0.02 L - 0.0249 L)
= 0.19502 mol/L / -0.0049 L
= - 39.79 mol/L (negative value because the volume of NaOH added is larger than the volume of HCl)
It is not physically possible to have a negative concentration, so the pH for this step of the titration cannot be calculated.
14. At the equivalence point, the moles of HCl and NaOH are stoichiometrically equivalent. This means that all the HCl has been neutralized by NaOH. At the equivalence point, the pH of the solution will be determined by the resulting salt. Since NaCl is formed from the neutralization reaction, which is a neutral compound, the pH at the equivalence point will be neutral, around 7.
15. After adding 25.1 mL of NaOH, we can follow the same steps as in step 13 to calculate the concentration of HCl remaining. However, the calculation will not be possible because the volume of NaOH added is larger than the volume of HCl initially.
16. The pH of the equivalence point during a certain titration is given as 3.5.
17. An indicator that changes color around pH 3.5 would be adequate for this titration. An example of such an indicator is phenolphthalein, which changes from colorless to pink around pH 8.2-10.0.
18. An indicator that has a color change around pH 3.5 would need to be avoided. One example of such an indicator is bromothymol blue, which has a color change around pH 6.0-7.6. Using this indicator would make it difficult to accurately determine the equivalence point at pH 3.5.
19. To calculate the volume of NaOH 0.5 mol/L necessary to neutralize 300 mL of HCl 0.2 M, we can use the fact that the moles of HCl and NaOH at the equivalence point are stoichiometrically equivalent.
Moles of HCl = Molarity * Volume
= 0.2 mol/L * 0.3 L
= 0.06 moles
Since NaOH reacts in a 1:1 ratio with HCl, the volume of NaOH needed can be calculated by dividing the moles of HCl by the molarity of NaOH.
Volume of NaOH = Moles of HCl / Molarity of NaOH
= 0.06 moles / 0.5 mol/L
= 0.12 L
= 120 mL
Therefore, 120 mL of NaOH 0.5 mol/L is necessary to neutralize 300 mL of HCl 0.2 M.
To calculate the pH of a solution during an acid-base titration, we need to consider the stoichiometry of the reaction and the concentration of the acid and the base. Let's go step by step through each question:
12. Before the titration:
At this point, there is no reaction between the acid and the base. Therefore, the pH of the solution is determined by the dissociation of the solvent, which is usually water. The pH of pure water is 7, indicating neutrality.
13. After adding 24.9 mL of NaOH:
Before calculating the pH, we need to determine the remaining concentration of HCl after reacting with 24.9 mL of NaOH. To do this, we can use the stoichiometry of the reaction. From the given information, we know that 25 mL of NaOH 0.2 M neutralizes 20 mL of HCl.
Let's calculate the concentration of HCl:
(20 mL HCl) x (0.2 mol/L NaOH) / (25 mL NaOH) = 0.16 M HCl
Now, we can calculate the pH of the solution considering the remaining concentration of HCl. We can use the Henderson-Hasselbalch equation, which is commonly used to calculate pH in buffer solutions:
pH = -log[H+]
pH = -log(0.16)
pH ≈ 0.80
So, the pH of the solution after adding 24.9 mL of NaOH is approximately 0.80.
14. At the equivalence point:
The equivalence point is reached when the stoichiometric amount of acid has reacted with the base. In this case, since NaOH and HCl react in a 1:1 ratio, the equivalence point is reached when 25 mL of NaOH have been added.
At this point, the solution will be neutral because all the acid has reacted with the base. Therefore, the pH of the solution at the equivalence point is 7 (neutral).
15. After adding 25.1 mL of NaOH:
After adding 25.1 mL of NaOH, we have slightly excess NaOH. We can calculate the concentration of excess NaOH by subtracting the amount of NaOH that reacted from the total amount added:
25.1 mL - 25 mL (equivalence point volume) = 0.1 mL
The pH of the solution will still remain close to 7 after adding 0.1 mL of excess NaOH, indicating a slightly basic solution.
16. If the pH at the equivalence point is 3.5:
At the equivalence point, the amount of acid will be equal to the amount of base, resulting in a neutral solution. However, a pH of 3.5 indicates an acidic solution. This suggests that an indicator with a color change in the acidic range could be used. Examples of indicators suitable for this pH range include bromothymol blue, bromocresol green, and methyl orange.
17. Adequate indicator for this titration:
Based on the pH at the equivalence point (3.5), an indicator that changes color in the acidic range is required. Bromothymol blue is one possible indicator that could be used in this titration since it changes color between pH 6 and 7. Other indicators suitable for this pH range, such as bromocresol green or methyl orange, could also be used.
18. Indicator to avoid during this titration:
An indicator to avoid would be one that changes color outside the pH range of the titration endpoint. For example, phenolphthalein changes color around pH 8-9, which is higher than the expected pH at the equivalence point (3.5). Therefore, phenolphthalein would not provide an accurate indication of the endpoint in this titration.
19. To calculate the volume of NaOH required to neutralize HCl:
We can use the stoichiometry of the reaction to calculate the volume of NaOH needed. From the given information, we know that HCl and NaOH react in a 1:1 ratio.
Let's calculate the moles of HCl:
(300 mL HCl) x (0.2 mol/L HCl) = 60 mmol HCl
Since the reaction is 1:1, we need the same amount of NaOH. We can calculate the volume of NaOH using its concentration:
(60 mmol NaOH) / (0.5 mol/L NaOH) = 120 mL NaOH
Therefore, 120 mL of NaOH 0.5 mol/L is necessary to neutralize 300 mL of HCl 0.2 M.