How many milliliters of 0.0050 N KOH are required to neutralize 19 mL of 0.0050 N H2SO4?
How many milliliters of 0.0050 N KOH are required to neutralize 19 mL of 0.0050 N HCl?
For #2, mL x M = mL x M and that works because HCl and NaOH react 1:1.
For the first one,
moles H2SO4 = M x L.
Convert moles H2SO4 to moles KOH using the coefficients in the balanced equation.
Then M KOH = moles KOH/L of solution.
To find the volume of 0.0050 N KOH required to neutralize 19 mL of 0.0050 N H2SO4 and 0.0050 N HCl, we can use the concept of stoichiometry and the equation:
H2SO4 + 2 KOH → K2SO4 + 2 H2O
HCl + KOH → KCl + H2O
For the first question, the equation is balanced and shows that 2 moles of KOH are required to neutralize 1 mole of H2SO4.
For the second question, the equation is also balanced and shows that 1 mole of KOH is required to neutralize 1 mole of HCl.
Using the equation, we can determine the moles of H2SO4 and HCl:
Moles of H2SO4 = Volume of H2SO4 (in L) x Concentration of H2SO4 (in mol/L) = 19 mL x 0.0050 N / 1000 mL/L = 0.000095 mol
Moles of HCl = Volume of HCl (in L) x Concentration of HCl (in mol/L) = 19 mL x 0.0050 N / 1000 mL/L = 0.000095 mol
For the first question:
To neutralize 0.000095 mol of H2SO4, we will require 2 x 0.000095 mol = 0.00019 mol of KOH.
Now, we can find the volume of 0.0050 N KOH required using the given concentration:
Volume of KOH (in L) = Moles of KOH / Concentration of KOH = 0.00019 mol / 0.0050 N = 0.038 L
Finally, we convert the volume to milliliters:
Volume of KOH (in mL) = Volume of KOH (in L) x 1000 mL/L = 38 mL
Therefore, 38 milliliters of 0.0050 N KOH are required to neutralize 19 mL of 0.0050 N H2SO4.
For the second question:
To neutralize 0.000095 mol of HCl, we will require 0.000095 mol of KOH.
Volume of KOH (in L) = Moles of KOH / Concentration of KOH = 0.000095 mol / 0.0050 N = 0.019 L
Volume of KOH (in mL) = Volume of KOH (in L) x 1000 mL/L = 19 mL
Therefore, 19 milliliters of 0.0050 N KOH are required to neutralize 19 mL of 0.0050 N HCl.
To find the number of milliliters of 0.0050 N KOH required to neutralize each solution, we can use the concept of stoichiometry and the balanced chemical equations.
First, let's write the balanced chemical equations for the neutralization reactions:
1) H2SO4 + 2 KOH → K2SO4 + 2 H2O
2) HCl + KOH → KCl + H2O
For equation 1, we can see that one mole of H2SO4 reacts with 2 moles of KOH. This means that the ratio of H2SO4 to KOH is 1:2.
Similarly, for equation 2, we can see that one mole of HCl reacts with 1 mole of KOH. This means that the ratio of HCl to KOH is 1:1.
Next, we can calculate the number of moles of H2SO4 and HCl in each solution:
Moles of solute = concentration (N) × volume (in liters)
For both solutions, the concentration is 0.0050 N and the volume is 19 mL, which is 0.019 L.
Moles of H2SO4 = 0.0050 N × 0.019 L = 9.5 × 10^(-5) moles
Moles of HCl = 0.0050 N × 0.019 L = 9.5 × 10^(-5) moles
Now, let's calculate the number of moles of KOH needed to neutralize each solution based on the stoichiometric ratios:
1) Moles of KOH needed = Moles of H2SO4 × (2 moles KOH / 1 mole H2SO4)
= 9.5 × 10^(-5) moles × (2 moles / 1 mole)
= 1.9 × 10^(-4) moles
2) Moles of KOH needed = Moles of HCl × (1 mole KOH / 1 mole HCl)
= 9.5 × 10^(-5) moles × (1 mole / 1 mole)
= 9.5 × 10^(-5) moles
Finally, to find the volume of 0.0050 N KOH needed, we can use the equation:
Volume (in mL) = Moles of solute / concentration (N) × 1000
1) Volume of 0.0050 N KOH for H2SO4 = (1.9 × 10^(-4) moles / 0.0050 N) × 1000
= 38 mL
2) Volume of 0.0050 N KOH for HCl = (9.5 × 10^(-5) moles / 0.0050 N) × 1000
= 19 mL
Therefore, 38 mL of 0.0050 N KOH is required to neutralize 19 mL of 0.0050 N H2SO4, and 19 mL of 0.0050 N KOH is required to neutralize 19 mL of 0.0050 N HCl.