I'm not really sure how to do this. I've tried many ways and I can't get the right answer of 0.0569 mol/L H3PO4
100.0 mL of a solution containing 2.55g of potassium hydroxide is mixed with 100.0 mL of a 1.50 mol/L phosphoric acid solution. Will the resulting mixture be acidic or basic? What will be the molarity of the potassium phosphate formed? (Assume the volumes are additive).
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My work:
3KOH + H3PO4 --> K3PO4 + 3 H2O
I found moles of K3PO4 by taking mass of KOH 2.55 g and dividing it by molar mass 56.11 g/mol, multiplying by mol ratio of 1 mol of K3PO4 over 3 mol of KOH. I got 0.0151488148 mol K3PO4
I found moles of K3PO4 by taking volume of H3PO4 0.1000 L and multiplying it by the concentration 1.50 mol/L, then multiplying by mol ratio of 1 mol K3PO4 over 1 mol of H3PO4. I got 0.015 mol K3PO4.
Now I have to find limiting reagent? But aren't they the same mol? Not sure what to do next. Please help
I don't see anything wrong with what you've done and I don't consider the two answers the same. To three significant figures 0.0151 vs 0.0150 so all of the H3PO4 (the limiting reagemt) is used and a slight excess of KOH is present. Note that the problem asks for M of K3PO4 and not H3PO4. I don't get 0.0569 M for K3PO4. I find M K3PO4 = mols K3PO4/L = 0.015/0.2 = ?
Okay thank you
To determine the limiting reagent, compare the number of moles of each reactant involved in the balanced equation. In this case, we have:
1. Moles of KOH: 2.55 g / 56.11 g/mol = 0.04545 mol KOH
2. Moles of H3PO4: 0.1 L * 1.50 mol/L = 0.15 mol H3PO4
Comparing the moles of the two reactants, we can see that the KOH is the limiting reagent because it has less moles compared to H3PO4.
To find the moles of K3PO4 formed, use the stoichiometry of the balanced equation:
1 mol KOH reacts with 1 mol K3PO4
Since KOH is the limiting reagent, we have:
0.04545 mol KOH * (1 mol K3PO4 / 3 mol KOH) = 0.01515 mol K3PO4
So the moles of K3PO4 formed is 0.01515 mol.
To find the molarity of the K3PO4 formed, divide the moles of K3PO4 by the total volume of the solution:
Total volume = 100 mL + 100 mL = 0.1 L + 0.1 L = 0.2 L
Molarity of K3PO4 = moles of K3PO4 / total volume
= 0.01515 mol / 0.2 L
= 0.07575 mol/L
So the molarity of the potassium phosphate formed is 0.07575 mol/L.
Since K3PO4 is a salt of a weak acid (H3PO4) and a strong base (KOH), it will have a slightly basic pH. Therefore, the resulting mixture will be slightly basic.
To determine the limiting reagent in this reaction, you need to compare the moles of KOH and H3PO4.
You have correctly calculated the moles of K3PO4 produced from both compounds separately: 0.0151488148 mol of K3PO4 from KOH and 0.015 mol of K3PO4 from H3PO4.
Since the mole ratios between KOH and K3PO4 are 3:1, you can see that the moles of K3PO4 produced from KOH (0.0151488148 mol) exceed the moles of K3PO4 produced from H3PO4 (0.015 mol), indicating that KOH is the limiting reagent in this reaction.
Therefore, you need to calculate the molarity of the resulting K3PO4 solution based on the moles of K3PO4 generated from the limiting reagent (KOH).
To find the molarity, divide the moles of K3PO4 by the total volume of the solution.
Total volume = volume of KOH solution + volume of H3PO4 solution = 100.0 mL + 100.0 mL = 200.0 mL = 0.2000 L
Molarity of K3PO4 = moles of K3PO4 / total volume of solution = (0.0151488148 mol K3PO4) / (0.2000 L) = 0.0757 mol/L
Therefore, the molarity of the resulting K3PO4 solution is 0.0757 mol/L.
Since K3PO4 is a salt formed by the reaction between a strong base (KOH) and a weak acid (H3PO4), the resulting mixture will be basic.