Determine the volume of 0.240 M KOH solution required to neutralize each of the following samples of sulfuric acid. The neutralization reaction is shown below.
H2SO4(aq) + 2 KOH(aq)--> K2SO4(aq) + 2 H2O(l)
(a) 26 mL of 0.240 M H2SO4
thanks
How many moles H2SO4 do you have. M x L = moles.
Using the coefficients in the balanced equation, convert moles H2SO4 to moles KOH. Note that 1 mole H2SO4 = 2 mole KOH.
Now M KOH = moles KOH/L KOH. Solve for L KOH.
To determine the volume of the 0.240 M KOH solution required to neutralize 26 mL of 0.240 M H2SO4, we can use the concept of stoichiometry and the balanced equation for the neutralization reaction.
First, let's identify the balanced equation:
H2SO4(aq) + 2 KOH(aq) → K2SO4(aq) + 2 H2O(l)
From the equation, we can see that the molar ratio between H2SO4 and KOH is 1:2. This means that for every one mole of H2SO4, we need 2 moles of KOH to completely neutralize it.
Next, let's calculate the number of moles of H2SO4 in the 26 mL of the solution using the formula:
moles = concentration × volume
moles of H2SO4 = 0.240 M × 0.026 L (since 26 mL = 0.026 L)
Now that we have the moles of H2SO4, we can use the ratio from the balanced equation to determine the number of moles of KOH required to neutralize it. Since the ratio is 1:2, we multiply the moles of H2SO4 by 2:
moles of KOH = 2 × moles of H2SO4
Finally, to calculate the volume of the 0.240 M KOH solution required, we can use the formula:
volume = moles / concentration
volume of KOH solution = moles of KOH / 0.240 M
Substituting the values into the equation, we can find the answer by performing the calculation.