What is the pH at the titration of 40 mL of 1.5 M Tartaric acid
with 30ml of 0.6 M NaOH?
To find the pH at the titration of Tartaric acid with NaOH, we need to determine the number of moles of acid and base involved in the reaction. Then we can calculate the concentration of the resulting solution and use this information to find the pH.
First, let's calculate the number of moles of tartaric acid and NaOH used in the reaction:
Moles of tartaric acid = volume (in L) * concentration (in mol/L)
= 0.040 L * 1.5 M
= 0.06 moles
Moles of NaOH = volume (in L) * concentration (in mol/L)
= 0.030 L * 0.6 M
= 0.018 moles
Next, we need to determine which reactant is in excess. To do that, we compare the moles of tartaric acid and NaOH used. Since tartaric acid has a higher number of moles (0.06 moles) compared to NaOH (0.018 moles), tartaric acid is in excess, and NaOH is the limiting reactant.
Now, let's determine the number of moles of tartaric acid that reacted with NaOH. Since the reaction ratio between tartaric acid and NaOH is 1:2, we can multiply the number of moles of NaOH used by 2:
Moles of tartaric acid reacted = 2 * moles of NaOH
= 2 * 0.018 moles
= 0.036 moles
To find the final volume of the solution after the reaction, we need to add the volumes of tartaric acid and NaOH used:
Final volume = volume of tartaric acid + volume of NaOH
= 0.040 L + 0.030 L
= 0.070 L
Next, let's calculate the concentration of the resulting solution:
Concentration (in mol/L) = moles of tartaric acid reacted / final volume (in L)
= 0.036 moles / 0.070 L
= 0.514 M
Now we can find the pOH of the resulting solution using the concentration we just calculated:
pOH = -log10(concentration)
= -log10(0.514)
≈ 0.29
Since pH + pOH = 14, we can find the pH:
pH = 14 - pOH
= 14 - 0.29
≈ 13.71
Therefore, the pH at the titration of 40 mL of 1.5 M tartaric acid with 30 mL of 0.6 M NaOH is approximately 13.71.