What volume of 0.076M HNO3 is required to titrate 20.0ML of 0.25M Ba(OH)2 solution? use 2 sig figures
Ba(OH)2 + 2HNO3 ==> 2H2O + Ba(NO3)2
mols Ba(OH)2 = M x L = ?
Using the coefficients in the balanced equation convert mols Ba(OH)2 to mols HNO3.
Then M HNO3 = mols HNO3/L HNO3. You have M and mols, solve for L. Convert to mL if desired.
To find the volume of 0.076 M HNO3 required to titrate 20.0 mL of 0.25 M Ba(OH)2 solution, we can use the concept of stoichiometry and the balanced chemical equation between HNO3 and Ba(OH)2.
The balanced chemical equation for the reaction between HNO3 and Ba(OH)2 is:
2 HNO3 + Ba(OH)2 → Ba(NO3)2 + 2 H2O
From the equation, we can see that 2 moles of HNO3 react with 1 mole of Ba(OH)2 to produce 1 mole of Ba(NO3)2 and 2 moles of water.
First, let's calculate the number of moles of Ba(OH)2 in the given solution:
Moles of Ba(OH)2 = (volume of solution in liters) × (molarity of Ba(OH)2)
= 0.020 L × 0.25 M
= 0.005 mol
According to the stoichiometry of the balanced chemical equation, 2 moles of HNO3 are required to react with 1 mole of Ba(OH)2. Therefore, the number of moles of HNO3 required can be calculated as follows:
Moles of HNO3 = 2 × Moles of Ba(OH)2
= 2 × 0.005 mol
= 0.010 mol
Now, we can find the volume of 0.076 M HNO3 solution needed to provide this amount of moles:
Volume of HNO3 = (moles of HNO3) / (molarity of HNO3)
= 0.010 mol / 0.076 M
= 0.132 L = 132 mL
Finally, we round the answer to two significant figures as specified:
Volume of 0.076 M HNO3 = 132 mL