In one trial of an investigation 50.0 mL of HCl(aq) of an unknown concentration is Tiresias with 0.10M NaOH (aq). During the titration the total volume of NaOH (aq) added and corresponding pH value of the reaction mixture are measured and recorded in the table below: 10mL-1.6pH 20.0mL-2.2pH 24.0mL-2.9pH 24.9mL-3.9pH 25.1mL-10.1pH 26.0mL-11.1pH 30.0mL-11.8pH.
Determine the total volume of NaOH(aq) added when the reaction mixture has a pH value of 7.0?
Write a balanced equation that represents this neutralization reaction?
The large JUMP in pH is between 24.9 and 25.1 so the equivalence point must be between those two values of the volume. If you take the average pH value that will be (3.9+10.1)/2 = 7.0 so just average the volumes. That will be (24.9+25.1)/2 = ??
NaOH + HCl ==> NaCl + H20.
To determine the total volume of NaOH(aq) added when the reaction mixture has a pH value of 7.0, we need to find the point in the titration where the pH value reaches 7.0.
From the given table, we can see that the pH value starts to increase rapidly after 24.9 mL of NaOH(aq) is added. At this point, the pH value is 3.9. As more NaOH(aq) is added, the pH value continues to increase until it reaches 7.0.
To estimate the total volume of NaOH(aq) added at pH 7.0, we can interpolate between the data points provided.
Using a linear approximation, we can calculate:
Change in volume NaOH(aq) = (Change in pH value) / (Slope of the pH change)
Slope of the pH change = (pH2 - pH1) / (Volume2 - Volume1) = (3.9 - 2.9) / (24.9 - 20.0) = 1 / 4.9 ≈ 0.2041
Change in volume NaOH(aq) = (7.0 - 3.9) / 0.2041 ≈ 15.16 mL
Therefore, the total volume of NaOH(aq) added when the reaction mixture has a pH value of 7.0 is approximately 24.9 mL + 15.16 mL ≈ 40.06 mL.
Now, let's write a balanced equation that represents this neutralization reaction:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
This equation shows that in the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH), they react to form sodium chloride (NaCl) and water (H2O).
To determine the total volume of NaOH(aq) added when the reaction mixture has a pH value of 7.0, we need to look for the point where the pH value reaches 7.0 in the data provided.
From the given table:
10 mL - 1.6 pH
20.0 mL - 2.2 pH
24.0 mL - 2.9 pH
24.9 mL - 3.9 pH
25.1 mL - 10.1 pH
26.0 mL - 11.1 pH
30.0 mL - 11.8 pH
We can see that the pH increases significantly between 24.9 mL and 25.1 mL, indicating that the neutralization point is likely close to that range. Additionally, the pH jumps from 3.9 to 10.1, which is significant.
To find the exact volume, we can use linear interpolation.
Using the formula for the equation of a line, we can determine the volume at pH 7.0:
(Volume - 25.1 mL) / (10.1 pH - 3.9 pH) = (7.0 pH - 3.9 pH) / (10.1 pH - 3.9 pH)
Simplifying the equation:
Volume - 25.1 mL = (7.0 - 3.9) / (10.1 - 3.9)
Volume - 25.1 mL = 3.1 / 6.2
Volume - 25.1 mL = 0.5
Volume = 25.1 mL + 0.5 mL
Volume = 25.6 mL
Therefore, the total volume of NaOH(aq) added when the reaction mixture has a pH value of 7.0 is 25.6 mL.
Now, let's write the balanced equation that represents this neutralization reaction.
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
In this reaction, hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH) to form sodium chloride (NaCl) and water (H2O).