In a titration of 35.00 mL of 0.737 M S, __________ mL of a 0.827 M KOH solution is required for neutralization.
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To find the volume of the KOH solution required for neutralization, we need to use the concept of stoichiometry.
In this titration, we are adding a solution of KOH (potassium hydroxide) to a solution of S (sulfur). The balanced chemical equation for the reaction between the two components is:
S + 2KOH → K2S + 2H2O
From the balanced equation, we can see that 1 mole of S reacts with 2 moles of KOH.
We are given the volume and concentration of the S solution (35.00 mL of 0.737 M). Using this information, we can calculate the number of moles of S present in the solution:
moles of S = volume of S solution (in L) × concentration of S solution (in mol/L)
= 35.00 mL ÷ 1000 mL/L × 0.737 mol/L
Now, we can use the stoichiometry of the balanced equation to determine the moles of KOH required for neutralization. Since the ratio of S to KOH in the balanced equation is 1:2, we can say:
moles of KOH = 2 × moles of S
Finally, we can calculate the volume of KOH solution required for neutralization using the concentration of the KOH solution (0.827 M) and the number of moles of KOH:
volume of KOH solution (in L) = moles of KOH / concentration of KOH solution (in mol/L)
By substituting the values into the equation, we can find the answer to the question.