In a titration of 35.00 mL of 0.737 M S, __________ mL of a 0.827 M KOH solution is required for neutralization.

Question

In a titration of 35.00 mL of 0.737 M S, __________ mL of a 0.827 M KOH solution is required for neutralization.

Answer 1

See your post above.

Answer 2

To determine the volume of 0.827 M KOH solution required for neutralization, we can use the equation:

Molarity of acid x Volume of acid = Molarity of base x Volume of base

Given:
Molarity of acid (S) = 0.737 M
Volume of acid (S) = 35.00 mL
Molarity of base (KOH) = 0.827 M
Volume of base (KOH) = ?

Using the equation, we can rearrange it to solve for the volume of base:

Volume of base (KOH) = (Molarity of acid x Volume of acid) / Molarity of base

Plugging in the values, we get:

Volume of base (KOH) = (0.737 M x 35.00 mL) / 0.827 M

Now, let's calculate the volume of the KOH solution required for neutralization.

Answer 3

To find the volume of the 0.827 M KOH solution required for neutralization, we can use the concept of stoichiometry. In this case, we need to determine the stoichiometric ratio between S and KOH in the balanced chemical equation for their reaction.

The balanced chemical equation for the reaction between S and KOH is:

S + 2KOH → K2S + H2O

From the equation, we see that 1 mole of S reacts with 2 moles of KOH. This means that the stoichiometric ratio between S and KOH is 1:2.

Now, let's calculate the amount of S used in the titration. We have 35.00 mL of a 0.737 M S solution. To get the amount in moles, we multiply the volume (in L) by the concentration (in mol/L):

Amount of S = Volume of S × Concentration of S
= 0.03500 L × 0.737 mol/L
= 0.025795 moles

According to the stoichiometric ratio, 1 mole of S reacts with 2 moles of KOH. Therefore, the amount of KOH needed can be calculated as follows:

Amount of KOH = Amount of S × Stoichiometric ratio
= 0.025795 moles × 2
= 0.05159 moles

Now, let's calculate the volume of the 0.827 M KOH solution required. We have the amount of KOH in moles, and we can use the concentration and the formula:

Volume of KOH = Amount of KOH ÷ Concentration of KOH
= 0.05159 moles ÷ 0.827 mol/L
≈ 0.0624 L or 62.4 mL

Therefore, about 62.4 mL of the 0.827 M KOH solution is required for neutralization in the titration.

In a titration of 35.00 mL of 0.737 M S, __________ mL of a 0.827 M KOH solution is required for neutralization.

62.4mL

See your post above.

To determine the volume of 0.827 M KOH solution required for neutralization, we can use the equation:

To find the volume of the 0.827 M KOH solution required for neutralization, we can use the concept of stoichiometry. In this case, we need to determine the stoichiometric ratio between S and KOH in the balanced chemical equation for their reaction.