I know the formula is Ma x Mla / 1000 x Sa / Sb = Mb x Mlb / 1000
if 175 ml of KOH of unknown concentration with a methyl red indicator present is titrated with [0.275 M] H3PO4, the yellow solution turns red after 29.6 ml of the H3PO4 is added. What is the molarity of the KOH solution?
On the above problem I think it goes like this but i'm stumped...
0.172 x 27.4 / 1000 x 1/2 = Mb x 100 / 1000 ?? What do I do with the 1/2
also..
If 100 ml of HCl of unknown concentration with a methyl orange indicator present is titrated with [0.172 M] Ca(OH)2, the red solution turns yellow after 27.4 ml of the Ca(OH)2 is added, What is the molarity of the HCl solution?
Please Help I can't understand how to do this...
moles H3PO4 = 0.0296 L x 0.275 M=0.00814
the balanced equation is
H3PO4 + 3 NaOH = Na3PO4 + 3 H2O
the ratio between H3PO4 and NaOH is 1 : 3
moles NaOH = 3 x 0.00814 =0.0244
molarity of NaOH = 0.0244 mol/ 0.175 L=0.140 M
To solve the first problem, which involves finding the molarity of the KOH solution, let's break down the steps:
1. First, let's write down the given information:
- Volume of KOH solution (Vla) = 175 ml
- Volume of H3PO4 solution added (Sa) = 29.6 ml
- Concentration of H3PO4 solution (Ca) = 0.275 M
2. Next, let's calculate the moles of H3PO4 used in the reaction using the formula:
Moles of H3PO4 = (Concentration of H3PO4) x (Volume of H3PO4 solution in liters)
= 0.275 M x (29.6 ml / 1000 ml/L)
= 0.00812 moles
3. The balanced chemical equation between KOH and H3PO4 is:
3 KOH + H3PO4 -> K3PO4 + 3 H2O
From the equation, we can see that the stoichiometric ratio between H3PO4 and KOH is 1:3. We need to use the stoichiometry to find the moles of KOH:
Moles of KOH = (Moles of H3PO4) x (Stoichiometric ratio of KOH to H3PO4)
= 0.00812 moles x (3 moles KOH / 1 mole H3PO4)
= 0.0244 moles KOH
4. Now, let's find the volume of the KOH solution in liters:
Volume of KOH solution (Vla) = 175 ml = 0.175 L
5. Finally, we can calculate the molarity of the KOH solution (Mb) using the formula:
Mb = (Moles of KOH) / (Volume of KOH solution in liters)
= 0.0244 moles / 0.175 L
≈ 0.14 M
Thus, the molarity of the KOH solution is approximately 0.14 M.
Regarding your concern about the "1/2" in your equation, it seems like it should not be there. The formula you provided initially does not have that term. Please recheck the provided information or equation to ensure accuracy.
Now, let's move on to the second problem involving finding the molarity of the HCl solution.
1. Write down the given information:
- Volume of HCl solution (Vla) = 100 ml
- Volume of Ca(OH)2 solution added (Sa) = 27.4 ml
- Concentration of Ca(OH)2 solution (Ca) = 0.172 M
2. Calculate the moles of Ca(OH)2 used in the reaction:
Moles of Ca(OH)2 = (Concentration of Ca(OH)2) x (Volume of Ca(OH)2 solution in liters)
= 0.172 M x (27.4 ml / 1000 ml/L)
= 0.00471 moles
3. The balanced chemical equation between HCl and Ca(OH)2 is:
2 HCl + Ca(OH)2 -> CaCl2 + 2 H2O
From the equation, we can see that the stoichiometric ratio between HCl and Ca(OH)2 is 2:1. We need to use this ratio to find the moles of HCl:
Moles of HCl = (Moles of Ca(OH)2) x (Stoichiometric ratio of HCl to Ca(OH)2)
= 0.00471 moles x (2 moles HCl / 1 mole Ca(OH)2)
= 0.00942 moles HCl
4. Find the volume of the HCl solution in liters:
Volume of HCl solution (Vla) = 100 ml = 0.1 L
5. Calculate the molarity of the HCl solution (Mb):
Mb = (Moles of HCl) / (Volume of HCl solution in liters)
= 0.00942 moles / 0.1 L
= 0.0942 M
Therefore, the molarity of the HCl solution is 0.0942 M.
Hopefully, these explanations will help you understand how to approach similar problems in the future.