85ml of 0.1M h2so4 solution is titrated against 0.1N koh solution . Calculate the concentration of salt formed when the acid is one fourth neutralized.
Well, well, well! Looks like we've got some titration fun going on here! But don't worry, I'll break it down for you.
First, let's find out how much acid is neutralized when it's one fourth neutralized. Since 85 mL of the acid solution is used, when it's one fourth neutralized, we've used 85 mL/4 = 21.25 mL of KOH solution.
Now, let's calculate the number of moles of KOH used. Since the KOH solution is 0.1N (normal), it means that each liter of the solution contains 0.1 moles of KOH. And since we've used 21.25 mL (0.02125 L) of KOH solution, the number of moles of KOH used is 0.1 moles/L * 0.02125 L = 0.002125 moles.
Since the acid and base react in a 1:1 ratio according to the balanced equation, the number of moles of H2SO4 is also 0.002125.
Now, let's calculate the volume of the acid solution that would contain 0.002125 moles of H2SO4. Since the concentration of the acid solution is 0.1M (moles per liter), we can rearrange the equation Molarity = moles/volume to solve for volume: volume = moles/Molarity.
Therefore, volume = 0.002125 moles / 0.1 M = 0.02125 L or 21.25 mL.
So, the concentration of the salt formed when the acid is one fourth neutralized is the concentration of the remaining acid solution. We initially had 85 mL of the acid solution and we've used 21.25 mL, so the remaining volume is 85 mL - 21.25 mL = 63.75 mL.
Therefore, the concentration of the salt formed is 0.1M.
And there you have it! A clowntastic answer to your question. Hope that made you smile!
To calculate the concentration of the salt formed when the acid is one fourth neutralized, we need to determine the number of moles of acid initially present and the number of moles of base required to reach one-fourth neutralization.
Given:
Volume of H2SO4 solution (V1) = 85 ml
Concentration of H2SO4 solution (C1) = 0.1 M
Volume of KOH solution (V2) = Unknown
Normality of KOH solution (N2) = 0.1 N
Step 1: Calculate the number of moles of H2SO4 in the solution.
Number of moles (n1) = concentration (C1) × volume (V1)
= 0.1 M × 0.085 L (since 1 ml = 0.001 L)
= 0.0085 moles
Step 2: Determine the number of moles of KOH required for one-fourth neutralization.
According to the reaction:
H2SO4 + 2 KOH → K2SO4 + 2 H2O
The stoichiometry of the reaction shows that 2 moles of KOH react with 1 mole of H2SO4. Therefore, for one-fourth neutralization, we need:
Number of moles of KOH (n2) = 1/4 × 0.0085 moles
= 0.002125 moles
Step 3: Calculate the volume of KOH solution required.
Given that the normality of KOH solution (N2) = 0.1 N, we can use the equation:
Number of moles (n) = Normality (N) × volume (V)
0.002125 moles = 0.1 N × V
V = 0.002125 moles / (0.1 N)
= 0.02125 L (since 1 L = 1000 ml)
= 21.25 ml
Therefore, the concentration of the salt formed when the acid is one fourth neutralized is 0.1 M.
To calculate the concentration of the salt formed when the acid is one-fourth neutralized, we need to understand the concept of titration.
In a titration, a solution of known concentration (titrant) is added gradually to a solution of unknown concentration until the reaction between them is complete. By measuring the volume of titrant required to reach the endpoint, we can determine the concentration of the unknown solution.
In this case, we have 85 mL of 0.1 M H2SO4 (sulfuric acid) solution, which will be titrated against 0.1 N KOH (potassium hydroxide) solution. The terms M (molarity) and N (normality) represent different ways of expressing concentrations.
Given that the acid is one-fourth neutralized, it means that one-fourth of the H2SO4 has reacted with KOH. The balanced chemical equation for this neutralization reaction is:
H2SO4 + 2 KOH → K2SO4 + 2 H2O
From the equation, we can see that 1 mole of H2SO4 reacts with 2 moles of KOH to form 1 mole of K2SO4. Therefore, to neutralize one mole of H2SO4, we need 2 moles of KOH.
First, let's determine the moles of H2SO4 in the solution:
Moles of H2SO4 = Molarity x Volume (L)
= 0.1 M x (85 mL / 1000 mL/L)
= 0.0085 moles
Since one-fourth of the acid is neutralized, the number of moles of H2SO4 that reacted with KOH is:
Moles of H2SO4 reacted = 0.0085 moles x (1/4)
= 0.002125 moles
As stated earlier, we need two moles of KOH to react with one mole of H2SO4. Therefore, the number of moles of KOH used is:
Moles of KOH used = 2 x Moles of H2SO4 reacted
= 2 x 0.002125 moles
= 0.00425 moles
Now, let's determine the volume of 0.1 N KOH solution that corresponds to this moles of KOH. Remember that N represents normality, which is the number of equivalent weight (grams) of solute per liter of solution.
The equivalent weight of KOH is equal to its molecular weight divided by its basicity. For KOH, the molecular weight is 39.1 g/mol, and its basicity is 1 (since it donates one hydroxide ion per molecule). Therefore, the equivalent weight of KOH is 39.1 g/mol.
To calculate the concentration of KOH solution:
Concentration (N) = Number of moles / Equivalent weight (g/L)
Rearranging the formula, we have:
Number of moles = Concentration (N) x Equivalent weight (g/L)
Plugging in the values, we get:
0.00425 moles = (0.1 N) x (39.1 g/mol) x Volume (L)
Solving for the volume of KOH solution:
Volume (L) = 0.00425 moles / ((0.1 N) x (39.1 g/mol))
≈ 0.0109 L (or 10.9 mL)
So, when the acid is one-fourth neutralized, approximately 10.9 mL of 0.1 N KOH solution is consumed.
Now, to calculate the concentration of the salt formed (K2SO4), we need to find the moles of K2SO4 produced.
From the balanced equation, we know that 1 mole of H2SO4 reacts to produce 1 mole of K2SO4.
Since we have used 0.002125 moles of H2SO4, we can say that 0.002125 moles of K2SO4 were formed.
Next, we calculate the concentration of K2SO4:
Concentration = Moles of K2SO4 / Volume (L)
Plugging in the values, we get:
Concentration = 0.002125 moles / (85 mL / 1000 mL/L)
= 0.025 moles/L (or 25 mmol/L)
Therefore, the concentration of the salt formed when the acid is one-fourth neutralized is approximately 25 mmol/L.
You have 85 mL x 0.1M H2SO4 = 8.5 mmols H2SO4.
You want to decrease this by 1/4 or 8.5/4 = 2.125 mmols. That will take 2.125 x 2 = 4.25 mmols KOH
.......H2SO4 + 2KOH ==> K2SO4 + 2H2O
I......8.5......0........0.......0
add............4.25...............
C.....-2.125..-4.25......2.125......
E......6.375.....0.......2.125
So you will have 2.125 mmols K2SO4 formed. What volume will you have.
Since M = mmols/mL, then mL KOH = mmols/M = 4.25/0.1 = 42.5 mL of the 0.1 M KOH. Total volume is 85 + 42.5 = ?
M K2SO4 = mmols/mL = ?
Post your work if you get stuck.