Question

The KOH concentration in aqueous solution is 4.39×10⁻⁴M. What is the solution pH? (S pts) 4. What is the 98% confidence interval for the data set Student's value for the following data set: 5.5, 6.0,5.0,6.5, and 7.0 cm.s=√p(x−x)/m-1π​​;μ=£±∗​□​ Report the shortest and longest lengths. (10pts) A solution is prepared by dissolving 21.42 g of anhydrous CaBr2​(FW 199.886) in enough water to make a total of 200.5 mL of solution. Calculate the molarity of CaBr2​. (5 pts)

Answer 1

The pH of a 4.39×10⁻⁴ M KOH solution is calculated by first determining the pOH and then subtracting it from 14. To find the molarity of a CaBr₂ solution, you divide the number of moles (calculated from mass and molar mass) by the volume in liters.

Step-by-step explanation:

To calculate the pH of the KOH solution with a concentration of 4.39×10⁻⁴ M, you need to recognize that KOH is a strong base and dissociates completely in water. Since there is one hydroxide ion per KOH molecule, [OH-] is equal to the concentration of KOH. You would then calculate the pOH by taking the negative logarithm of the hydroxide ion concentration:

pOH = -log([OH-])

pOH = -log(4.39×10⁻⁴)

Next, to find the pH, you subtract the pOH from 14:

pH = 14 - pOH

Calculating the molarity of CaBr₂ solution involves dividing the number of moles of CaBr₂ by the total volume of the solution in liters. You first calculate the number of moles of CaBr₂ by using its molar mass:

moles of CaBr₂ = mass (in grams) / molar mass

moles of CaBr₂ = 21.42 g / 199.886 g/mol

Then divide the moles by the volume in liters to find molarity:

Molarity = moles / volume (L)

Molarity = moles of CaBr₂ / 0.2005 L