What volume of each of the following bases will react completely with 25.00 mL of 0.200 M HCl?
a) 0.100 M NaOH
b)0.0500 M Sr(OH)2
c)0.250 M KOH
please explain and show detail I'm so lost on this problem... thanks!
To find the volume of each base that will react completely with 25.00 mL of 0.200 M HCl, we need to use the concept of stoichiometry and the balanced chemical equations for the reactions between the bases and HCl. Let's go through each part step by step:
a) 0.100 M NaOH:
1. Write the balanced chemical equation for the reaction between NaOH and HCl:
NaOH + HCl -> NaCl + H2O
2. Determine the stoichiometry of the reaction. From the equation, we can see that 1 mole of NaOH reacts with 1 mole of HCl.
3. Calculate the moles of HCl used:
Moles of HCl = Volume of HCl (in L) × Concentration of HCl (in mol/L)
Moles of HCl = 25.00 mL / 1000 mL/L × 0.200 mol/L = 0.00500 mol
4. Since the ratio between NaOH and HCl is 1:1, the moles of NaOH required will be the same as the moles of HCl used, which is 0.00500 mol.
5. Calculate the required volume of 0.100 M NaOH:
Volume of NaOH (in L) = Moles of NaOH / Concentration of NaOH (in mol/L)
Volume of NaOH = 0.00500 mol / 0.100 mol/L = 0.0500 L = 50.0 mL
Therefore, 50.0 mL of 0.100 M NaOH will react completely with 25.00 mL of 0.200 M HCl.
b) 0.0500 M Sr(OH)2:
1. Write the balanced chemical equation for the reaction between Sr(OH)2 and HCl:
Sr(OH)2 + 2HCl -> SrCl2 + 2H2O
2. Determine the stoichiometry of the reaction. From the equation, we can see that 1 mole of Sr(OH)2 reacts with 2 moles of HCl.
3. Calculate the moles of HCl used, as we did in part a:
Moles of HCl = 0.00500 mol
4. Since the ratio between Sr(OH)2 and HCl is 1:2, the moles of Sr(OH)2 required will be half of the moles of HCl used, which is 0.00500 mol/2 = 0.00250 mol.
5. Calculate the required volume of 0.0500 M Sr(OH)2:
Volume of Sr(OH)2 (in L) = Moles of Sr(OH)2 / Concentration of Sr(OH)2 (in mol/L)
Volume of Sr(OH)2 = 0.00250 mol / 0.0500 mol/L = 0.0500 L = 50.0 mL
Therefore, 50.0 mL of 0.0500 M Sr(OH)2 will react completely with 25.00 mL of 0.200 M HCl.
c) 0.250 M KOH:
1. Write the balanced chemical equation for the reaction between KOH and HCl:
KOH + HCl -> KCl + H2O
2. Determine the stoichiometry of the reaction. From the equation, we can see that 1 mole of KOH reacts with 1 mole of HCl.
3. Calculate the moles of HCl used, as we did in part a:
Moles of HCl = 0.00500 mol
4. Since the ratio between KOH and HCl is 1:1, the moles of KOH required will be the same as the moles of HCl used, which is 0.00500 mol.
5. Calculate the required volume of 0.250 M KOH:
Volume of KOH (in L) = Moles of KOH / Concentration of KOH (in mol/L)
Volume of KOH = 0.00500 mol / 0.250 mol/L = 0.0200 L = 20.0 mL
Therefore, 20.0 mL of 0.250 M KOH will react completely with 25.00 mL of 0.200 M HCl.