How many millimeters of a 15.0%, by mass solution of KOH(aq)(d = 1.14 g/ml ) are required to produce 26.0 L of a solution with pH = 11.40?
A 15% solution is 0.15 g KOH/g solution. And since the density is 1.14 g/mL, that is 0.15g/1.14 mL or about 0.13 g/1 mL. How much of that do you need?
pH = 11.4. Convert to pOH and from there to (OH^-). I get close to 2.5 M. That is 2.5 mols KOH/liter of solution. How much do you need? You want to make up 26.0 L therefore, that is 2.5 moles KOH x 26.0 L = approximately 0.06 mols and that x molar mass KOH = about 3 grams KOH to make up to 26.0 L.
So set up a proportion
0.13 g KOH/1 mL KOH = 3.0 g KOH/? mL)
I get something like 26 mL or so. You need to go through and use more exact figures since I estimated here and there. Check my work.
I thought for the density, the calculation would be as follows: 0.15gKOH/g---> 1g of solution, density=1.14 g/ml.......0.15g/(1g/1.14g/ml) to get 0.171 g/ml. and then I thought molarity for [OH^-] was 2.51 * 10^-3. The proper set up I got like 3.66 g/? mL. I might be wrong!! plzz can u check what I did and have a final say on the answer
To solve this problem, you need to consider the molarity of the KOH(aq) solution and its volume. Here's how you can find the answer step-by-step:
Step 1: Determine the molarity of the KOH(aq) solution.
The pH of a solution can be used to calculate its hydroxide ion (OH-) concentration. In this case, the solution has a pH of 11.40, which means it is basic. To find the OH- concentration, you can use the formula: [OH-] = 10^(-pOH), where pOH = 14 - pH. Thus, pOH = 14 - 11.40 = 2.6.
Now, convert the pOH to [OH-] concentration: [OH-] = 10^(-2.6) = 0.0039811 M.
Step 2: Convert the [OH-] concentration to the molarity of the KOH solution.
In a 15.0% by mass solution of KOH(aq), 15.0 g KOH is dissolved in 100.0 g solution. Since the solution's density is given as 1.14 g/ml, this means that 100.0 g of the solution has a volume of 100.0/1.14 = 87.7193 mL.
To calculate the molarity, you need to find the moles of KOH in the solution. The molar mass of KOH is:
K = 39.10 g/mol
O = 16.00 g/mol
H = 1.01 g/mol
Adding these together, you get: 39.10 + 16.00 + 1.01 = 56.11 g/mol.
So in 15.0 g of KOH, there are 15.0 g / 56.11 g/mol = 0.2670 mol of KOH.
Now, divide the moles of KOH by the volume of the solution and convert mL to L: 0.2670 mol / 87.7193 mL * (1 L / 1000 mL) = 0.00304 M.
Step 3: Calculate the volume of the KOH solution needed.
To find the volume of the 0.00304 M KOH solution needed to produce 26.0 L of solution, you can use the equation: M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.
For this problem, M1 = 0.00304 M, V1 is what we need to find, M2 = 0.0039811 M, and V2 = 26.0 L.
Rearranging the equation, V1 = (M2 * V2) / M1 = (0.0039811 M * 26.0 L) / 0.00304 M = 34.2 L.
Step 4: Convert the volume to milliliters.
The solution requires 34.2 L of the 0.00304 M KOH solution. To convert this to milliliters, multiply by 1000: 34.2 L * 1000 mL/L = 34200 mL.
Therefore, you would need 34200 milliliters of the 15.0% by mass solution of KOH(aq) to produce 26.0 L of a solution with pH 11.40.