yes,it is soluble,as it makes sodium salt with NaOH.
-log(10^-4 M NaOH) = 4 14 - 4 = 10 pH NaOH -----------------------
The pH of a 1.0 x 10^-6 M NaOH solution is approximately 11.00. This is because NaOH is a strong base that dissociates completely in solution to produce hydroxide ions, resulting in a high pH.
H3O+ concentration in a 0.048 M NaOH solution is 2.4 x 10^-12 M. This is because NaOH is a strong base that dissociates completely in water to produce Na+ and OH- ions, which react with any H3O+ ions to form water. As a result, the H3O+ concentration in such a solution is extremely low.
To make a 0.2 M solution of NaOH, you would need to dissolve 1 mole of NaOH in enough water to make a final volume of 1 liter. This corresponds to approximately 40 grams of NaOH.
Molarity = moles of solute/Liters of solution ( 24.5 mL = 0.0245 L)Rearranged,moles of solute = Liters of solution * MolarityMoles NaOH = (0.0245 L)(0.130 M NaOH)= 3.19 X 10 -3 moles NaOH==================
-log(10^-4 M NaOH) = 4 14 - 4 = 10 pH NaOH -----------------------
The pH of a 1.0 x 10^-6 M NaOH solution is approximately 11.00. This is because NaOH is a strong base that dissociates completely in solution to produce hydroxide ions, resulting in a high pH.
H3O+ concentration in a 0.048 M NaOH solution is 2.4 x 10^-12 M. This is because NaOH is a strong base that dissociates completely in water to produce Na+ and OH- ions, which react with any H3O+ ions to form water. As a result, the H3O+ concentration in such a solution is extremely low.
Molarity = moles of solute/Liters of solution 3.42 M NaOH = 1.3 moles NaOH/Liters NaOH Liters NaOH = 1.3 moles NaOH/3.42 M NaOH = 0.38 Liters
To make a 0.2 M solution of NaOH, you would need to dissolve 1 mole of NaOH in enough water to make a final volume of 1 liter. This corresponds to approximately 40 grams of NaOH.
Balanced equation first.NaOH + HCl -> NaCl + H2OAll one to one so a simple equality will do here.(X ml)(0.43 M NaOH) = (10 ml)(0.1 M HCl)0.43X = 1X = 2.3 milliliters NaOH needed--------------------------------------
Molarity = moles of solute/Liters of solution ( 24.5 mL = 0.0245 L)Rearranged,moles of solute = Liters of solution * MolarityMoles NaOH = (0.0245 L)(0.130 M NaOH)= 3.19 X 10 -3 moles NaOH==================
To neutralize HCl with NaOH, the mole ratio is 1:1. So, the moles of HCl are 0.200 M x 0.020 L = 0.004 moles. Since NaOH and HCl react in a 1:1 ratio, we need 0.004 moles of NaOH. Using the molarity formula, we find that we need 0.010 L or 10.00 mL of 0.400 M NaOH.
The concentration is 12,8 g/L NaOH.
To find the molarity, you need to know the amount in moles of NaOH and the volume in liters. First, convert 10 mL to liters by dividing by 1000 (10 mL = 0.01 L). Then, calculate the number of moles of NaOH using the molarity formula, Molarity = moles/volume. Given that you have 0.05 moles of NaOH and a volume of 0.01 L, the molarity would be 5 M.
To prepare a 0.10 M NaOH solution, you need to dilute the 6.0 M solution. You can use the formula: M1V1 = M2V2 where M1 = 6.0 M, V1 is the volume of 6.0 M NaOH needed, M2 = 0.10 M, and V2 = 1000 mL (1.0 L). By substituting the values into the formula, you can calculate the volume of the 6.0 M NaOH needed to prepare the 0.10 M solution.
This solution contain 26,3 g NaOH.