What volume of each of the following bases will react completely with 21.00 mL of 0.500 M HCl?
(a) 0.100 M NaOH
(b) 0.0500 M Ba(OH)2
(c) 0.250 M KOH
moles acid or base = M x L = ??
Next use the coefficients in the balanced equation to convert moles acid/base to the other one. The M = moles/L. Here is an example using (a).
HCl + NaOH ==> NaCl + HOH
moles HCl = M x L = 0.500M x 0.021L = 0.0105.
moles NaOH = 0.0105 x (1 mole NaOH/1 mole HCl) = 0.0105 moles NaOH
M NaOH = moles NaOH/L NaOH. You know M NaOH and moles NaOH, solve for L NaOH and convert to mL.
Well, let's see if I can calculate this with a side of humor!
(a) 0.100 M NaOH: This one is like a tiny mouse trying to take on an elephant. The mouse may be small, but it's determined! To calculate the volume of NaOH needed, we can use the equation M1V1 = M2V2. Since we know the concentration of NaOH (0.100 M) and the volume of HCl (21.00 mL), we can rearrange the equation to solve for V2: V2 = (M1V1) / M2. Plugging in the values, the volume of NaOH needed would be [0.500 M * 21.00 mL] / 0.100 M.
(b) 0.0500 M Ba(OH)2: Ba(OH)2 may sound like a fancy abbreviation, but this barium is ready to party! Similar to the first question, we'll use the same equation, M1V1 = M2V2. Now, we just need to plug in the values for Ba(OH)2: [0.500 M * 21.00 mL] / 0.0500 M.
(c) 0.250 M KOH: Ah, the potent KOH. This one packs a punch! Time to use the equation again, M1V1 = M2V2. Let's calculate the volume of KOH needed: [0.500 M * 21.00 mL] / 0.250 M.
Remember, these calculations will give you the volumes of each base needed to react completely with the given amount of HCl. Just be careful not to spill anything, especially when dancing with Barium (Ba)!
Disclaimer: If you actually spill any of these chemicals, be sure to handle them safely and follow proper lab procedures. Safety first, humor second!
To determine the volume of each base that will react completely with the given amount of HCl, we can use the stoichiometry of the balanced chemical equations for the reactions between HCl and each of the bases.
(a) Reaction between HCl and NaOH:
HCl + NaOH → NaCl + H2O
The balanced chemical equation shows that one mole of HCl reacts with one mole of NaOH. Therefore, the stoichiometric ratio of HCl to NaOH is 1:1.
First, let's calculate the moles of HCl:
Moles of HCl = concentration × volume
= 0.500 M × 21.00 mL
= 0.500 mol/L × 0.02100 L
= 0.01050 mol
Since the stoichiometric ratio between HCl and NaOH is 1:1, the moles of NaOH required to react with the given amount of HCl is also 0.01050 mol.
To find the volume of 0.100 M NaOH that contains 0.01050 moles of NaOH, we can use the formula:
Volume (L) = Moles / Concentration
Volume of 0.100 M NaOH = 0.01050 mol / 0.100 mol/L
= 0.105 L = 105.0 mL
Therefore, the volume of 0.100 M NaOH required to react completely with 21.00 mL of 0.500 M HCl is 105.0 mL.
(b) Reaction between HCl and Ba(OH)2:
HCl + Ba(OH)2 → BaCl2 + 2H2O
The balanced chemical equation shows that two moles of HCl react with one mole of Ba(OH)2. Therefore, the stoichiometric ratio of HCl to Ba(OH)2 is 2:1.
We have already calculated that the moles of HCl is 0.01050 mol.
To find the moles of Ba(OH)2 required to react with 0.01050 mol HCl, we can use the stoichiometric ratio:
Moles of Ba(OH)2 = (2 moles of Ba(OH)2 / 1 mole of HCl) × 0.01050 mol
= 0.0210 mol
To find the volume of 0.0500 M Ba(OH)2 that contains 0.0210 moles of Ba(OH)2, we can use the formula:
Volume (L) = Moles / Concentration
Volume of 0.0500 M Ba(OH)2 = 0.0210 mol / 0.0500 mol/L
= 0.420 L = 420 mL
Therefore, the volume of 0.0500 M Ba(OH)2 required to react completely with 21.00 mL of 0.500 M HCl is 420 mL.
(c) Reaction between HCl and KOH:
HCl + KOH → KCl + H2O
The balanced chemical equation shows that one mole of HCl reacts with one mole of KOH. Therefore, the stoichiometric ratio of HCl to KOH is 1:1.
We have already calculated that the moles of HCl is 0.01050 mol.
To find the moles of KOH required to react with 0.01050 mol HCl, we can use the stoichiometric ratio:
Moles of KOH = moles of HCl = 0.01050 mol
To find the volume of 0.250 M KOH that contains 0.01050 moles of KOH, we can use the formula:
Volume (L) = Moles / Concentration
Volume of 0.250 M KOH = 0.01050 mol / 0.250 mol/L
= 0.0420 L = 42 mL
Therefore, the volume of 0.250 M KOH required to react completely with 21.00 mL of 0.500 M HCl is 42 mL.
To find the volume of each base that will react completely with 21.00 mL of 0.500 M HCl, we can use the concept of stoichiometry and the balanced chemical equation of the reaction.
First, let's write the balanced equation for the reaction between HCl and each of the bases:
(a) HCl + NaOH -> NaCl + H2O
(b) 2HCl + Ba(OH)2 -> BaCl2 + 2H2O
(c) HCl + KOH -> KCl + H2O
Now, let's calculate the amount (in moles) of HCl present in 21.00 mL of 0.500 M HCl.
Amount of HCl = concentration of HCl × volume of HCl = 0.500 mol/L × 0.02100 L = 0.0105 moles
(a) For the reaction between HCl and NaOH, the molar ratio is 1:1, meaning that 1 mole of HCl reacts with 1 mole of NaOH. Therefore, the amount (in moles) of NaOH needed to react completely with 0.0105 moles of HCl is also 0.0105 moles.
To find the volume of 0.100 M NaOH that corresponds to 0.0105 moles, we can use the equation:
Volume of NaOH = amount of NaOH / concentration of NaOH = 0.0105 moles / 0.100 mol/L = 0.105 L = 105.0 mL
Therefore, 105.0 mL of 0.100 M NaOH will react completely with 21.00 mL of 0.500 M HCl.
(b) For the reaction between HCl and Ba(OH)2, the molar ratio is 2:1, meaning that 2 moles of HCl react with 1 mole of Ba(OH)2. Therefore, the amount (in moles) of Ba(OH)2 needed to react completely with 0.0105 moles of HCl is 0.0105 / 2 = 0.00525 moles.
To find the volume of 0.0500 M Ba(OH)2 that corresponds to 0.00525 moles, we can use the equation:
Volume of Ba(OH)2 = amount of Ba(OH)2 / concentration of Ba(OH)2 = 0.00525 moles / 0.0500 mol/L = 0.105 L = 105.0 mL
Therefore, 105.0 mL of 0.0500 M Ba(OH)2 will react completely with 21.00 mL of 0.500 M HCl.
(c) For the reaction between HCl and KOH, the molar ratio is 1:1, meaning that 1 mole of HCl reacts with 1 mole of KOH. Therefore, the amount (in moles) of KOH needed to react completely with 0.0105 moles of HCl is also 0.0105 moles.
To find the volume of 0.250 M KOH that corresponds to 0.0105 moles, we can use the equation:
Volume of KOH = amount of KOH / concentration of KOH = 0.0105 moles / 0.250 mol/L = 0.042 L = 42.0 mL
Therefore, 42.0 mL of 0.250 M KOH will react completely with 21.00 mL of 0.500 M HCl.