To determine the pH of a solution of potassium hydroxide (KOH), we need to calculate the concentration of hydroxide ions (OH-) first. Since KOH dissociates completely in water, the concentration of OH- ions will be equal to the concentration of KOH. Therefore, the concentration of OH- ions in a 0.04 M KOH solution is also 0.04 M. To find the pOH, we take the negative logarithm (base 10) of the hydroxide ion concentration: pOH = -log(0.04) ≈ 1.4. Finally, to find the pH, we subtract the pOH from 14 (the sum of pH and pOH in water at 25°C): pH = 14 - 1.4 ≈ 12.6.
-log(1.0 X 10^-4 M KOH) = 4 14 - 4 = 10 pH KOH ----------------
The pH of a 1 x 10^-5 M KOH solution would be approximately 9. For a strong base like KOH, the pH can be calculated by taking the negative logarithm of the concentration of hydroxide ions.
The pH of a 10^-5 M KOH solution would be around 9. For a given concentration of a strong base like KOH, the pH can be calculated using the formula pH = 14 - pOH. Given that pOH = -log[OH-] and [OH-] = 10^-5 M in this case, pOH = 5. Therefore, pH = 14 - 5 = 9.
Molarity is the concentration of a solution, defined as moles per unit volume. Where, Molarity = moles / volume In this case the molarity of the HCl solution is 0.03 M The pH of this is calculated by the equation below pH = - log [H+] Where [H+] is the concentration of hydrogen ions/ protons present in the solution. As HCl only contains one hydrogen ion per molecule then the concentation of [H+] is 0.03 M Then the equation can be calculated (the minus sign is very important!) pH = - log 0.03 pH = 1.52 To summarise, Molarity of 0.03 M HCl solution is 0.03 M pH of 0.03 M HCl solution is 1.52
Molarity = moles of solute/Liters of solution get moles KOH 6.31 grams KOH (1 mole KOH/56.108 grams) = 0.11246 moles KOH 0.250 M KOH = 0.11246 moles KOH/XL 0.11246/0.250 = 0.4498 liters = 450 milliliters
-log(1.0 X 10^-4 M KOH) = 4 14 - 4 = 10 pH KOH ----------------
The pH of a 1 x 10^-5 M KOH solution would be approximately 9. For a strong base like KOH, the pH can be calculated by taking the negative logarithm of the concentration of hydroxide ions.
The pH of a 10^-5 M KOH solution would be around 9. For a given concentration of a strong base like KOH, the pH can be calculated using the formula pH = 14 - pOH. Given that pOH = -log[OH-] and [OH-] = 10^-5 M in this case, pOH = 5. Therefore, pH = 14 - 5 = 9.
-log(3.5 X 10^-4 M) = 3.4559 14 - 3.4559 = 10.5 pH
Molarity is the concentration of a solution, defined as moles per unit volume. Where, Molarity = moles / volume In this case the molarity of the HCl solution is 0.03 M The pH of this is calculated by the equation below pH = - log [H+] Where [H+] is the concentration of hydrogen ions/ protons present in the solution. As HCl only contains one hydrogen ion per molecule then the concentation of [H+] is 0.03 M Then the equation can be calculated (the minus sign is very important!) pH = - log 0.03 pH = 1.52 To summarise, Molarity of 0.03 M HCl solution is 0.03 M pH of 0.03 M HCl solution is 1.52
The pH of a 0.0670 M KOH solution can be calculated using the formula pH = 14 - pOH. Since KOH dissociates completely in water to produce OH- ions, the pOH can be found by taking the negative logarithm of the hydroxide ion concentration (0.0670 M in this case). Then, pH = 14 - pOH, allowing you to determine the solution's pH.
Molarity = moles of solute/Liters of solution get moles KOH 6.31 grams KOH (1 mole KOH/56.108 grams) = 0.11246 moles KOH 0.250 M KOH = 0.11246 moles KOH/XL 0.11246/0.250 = 0.4498 liters = 450 milliliters
To determine the number of moles of KOH in the solution, you can use the formula: moles = molarity x volume (L) First, convert the volume from mL to liters by dividing 750 mL by 1000. Then, multiply the molarity (5.00 M) by the volume in liters to find the number of moles of KOH in the solution.
To solve this problem, you first need to determine the moles of KOH present in the 30.0 mL sample. Then calculate the moles of HClO4 added after 39.9 mL. Based on these concentrations, determine the excess and limiting reagents to find the resulting pH. Consider the reaction that occurs between KOH and HClO4, and use the stoichiometry to calculate the amount of products formed. Finally, calculate the pH using the concentration of the resulting solution.
[OH-] = 1x10^-3 M[H+][OH-] = 1x10^-14[H+] = 1x10^-14/1x10^-3 = 1x10^-11pH = -log 1x10^-11 = 11Done another way:pOH = -log [OH-] = -log 1x10^-3 = 3pH + pOH = 14pH = 14 - 3 = 11
To find the molarity (M) of the KOH solution, first calculate the number of moles of KOH. The molar mass of KOH is approximately 56.11 g/mol. Therefore, 28 g of KOH corresponds to about 0.498 moles (28 g ÷ 56.11 g/mol). Finally, the molarity is calculated as moles of solute per liter of solution: ( M = \frac{0.498 \text{ moles}}{2 \text{ L}} = 0.249 , M ). Thus, the molarity of the solution is approximately 0.25 M.
The pH of a solution containing 0.1 M of HC2H3O2 is around 2.88.