pH = -log10[H] so 1.96 x 10-2 = -1.71 so 1.71.
pH + pOH = 14 14 - 1.71 = 12.29
To prepare a 5 M NaOH solution, measure out 200 g of NaOH pellets and dissolve them in enough water to make 1 liter of solution. To adjust the pH to 12, you can titrate the solution with a standard acid solution (e.g. HCl) until the desired pH is reached. Remember to wear appropriate safety gear and use a fume hood when working with NaOH.
The solubility increase from 38,7 g KCl/100g water to 40,7 g KCl/100 g water.
A 25 millimole (m mol) solution means there are 25 millimoles of solute in every liter of solution. It is a unit used to express the concentration of a solute in a solution.
Room temperature, typically around 25°C, can be converted to kelvins using the formula K = °C + 273.15. Therefore, 25°C is equivalent to 25 + 273.15 = 298.15 K. Thus, room temperature in kelvins is approximately 298.15 K.
About 80ml of water must be added to 40ml of a 25 percent by weight solution to make a 2 percent by weight solution.
The molarity of the solution is 0.5 M.
To prepare a 5 M NaOH solution, measure out 200 g of NaOH pellets and dissolve them in enough water to make 1 liter of solution. To adjust the pH to 12, you can titrate the solution with a standard acid solution (e.g. HCl) until the desired pH is reached. Remember to wear appropriate safety gear and use a fume hood when working with NaOH.
It would be approx 9042 litres.
The Ka value is a measure of the strength of an acid in solution, not a base like NaOH. Instead, the strength of a base is typically measured using the Kb value. However, if you are asking about the dissociation constant of NaOH in water, it would be the Kw value, which is equal to 1.0 x 10^-14 at 25 degrees Celsius.
No. of mol of NaOH used = 30/1000 x 0.125 = 0.00375No. of mol of HCl used = 25/1000 x 0.15 = 0.00375No. of mol of HCl used = No. of mol of NaOH used.Hence, the resultant solution will be neutral. pH ~ 7.(NaCl formed is a neutral salt, does not undergo any form of salt hydrolysis)
Well, isn't that a happy little question! To prepare 0.25 L of 0.5 M NaOH solution, you'll need to dissolve the correct amount of NaOH in water. You can use the formula C1V1 = C2V2 to calculate the amount of NaOH needed. Just take your time, follow the steps gently, and you'll have a perfect solution in no time.
Assuming that the 25.0 ml are added to the 475 ml of water, not diluted to that volume:Moles of NaOH in the original 25 ml = 25/1000 x 0.6 = 0.015 moles Final volume = 25 + 475 ml = 500 ml. We have 0.015 moles in 500 ml, so 0.03 moles in a litre, Molarity = 0.03.
The solubility increase from 38,7 g KCl/100g water to 40,7 g KCl/100 g water.
A 25 millimole (m mol) solution means there are 25 millimoles of solute in every liter of solution. It is a unit used to express the concentration of a solute in a solution.
[ x - 25 ] has no solution.
[ x - 25 ] has no solution.
First, since NaOH is a base you have to find the pOH first so you use the equation -> pOH = -log[NaOH] pOH = -log[NaOH] = -log[0.0111] pOH = 1.955 Then you use this equation -> 14 = pH + pOH to find the pH 14 = pH + pOH pH = 14 - pOH = 14 - 1.955 pH = 12.045 and that makes it basic Hope that helped. ^_^