Can use this way.
-log(0.001) = 3
14 - 3
= a pH of 11.0
The pH of a 0.001N NaOH solution is around 11.9. NaOH is a strong base, and at this concentration, it will result in a highly alkaline solution.
The amount of NaOH needed to raise the pH from 8 to 10 depends heavily on the conditions. The amount of NaOH needed will increase as the volume of the solution increases. Even more importantly, buffers can stabilize the pH significantly. If buffers are presently, the pH change will be much more gradual, and more NaOH will be required.
To calculate the pH of the resulting solution when mixing HCl and NaOH, you need to determine the moles of each reactant. Use these values to find the limiting reagent and the excess reagent. Then, calculate the concentration of the excess reagent left in solution. Finally, use the concentrations of the excess reagent and water to determine the pH of the resulting solution.
To calculate the pH of the solution after mixing, first find the moles of acetic acid (HOAc) and NaOH. Then, determine which reactant is limiting to find the excess OH-. Use the excess OH- to calculate the concentration of OH- in the final solution, and then use this to calculate the pH.
The pH of a 2.34x10^-5 NaOH solution is 12.33 (calculated as -log[OH^-]). The pOH of the same solution is 1.67 (calculated as -log[NaOH]).
The pH of a 0.001N NaOH solution is around 11.9. NaOH is a strong base, and at this concentration, it will result in a highly alkaline solution.
To calculate the pH after each volume of NaOH is added, you need to determine the moles of HBr initially, moles of NaOH added, and then use these values to calculate the final concentration of H3O+ ions in the solution. From there, you can calculate the pH using the equation pH = -log[H3O+]. Repeat these calculations for each volume of NaOH added to find the pH at 12.5 ml, 18.7 ml, 20.0 ml, 29.5 ml, and 34.2 ml.
The amount of NaOH needed to raise the pH from 8 to 10 depends heavily on the conditions. The amount of NaOH needed will increase as the volume of the solution increases. Even more importantly, buffers can stabilize the pH significantly. If buffers are presently, the pH change will be much more gradual, and more NaOH will be required.
To calculate the pH of the resulting solution when mixing HCl and NaOH, you need to determine the moles of each reactant. Use these values to find the limiting reagent and the excess reagent. Then, calculate the concentration of the excess reagent left in solution. Finally, use the concentrations of the excess reagent and water to determine the pH of the resulting solution.
To calculate the pH of the solution after mixing, first find the moles of acetic acid (HOAc) and NaOH. Then, determine which reactant is limiting to find the excess OH-. Use the excess OH- to calculate the concentration of OH- in the final solution, and then use this to calculate the pH.
The pH of a 2.34x10^-5 NaOH solution is 12.33 (calculated as -log[OH^-]). The pOH of the same solution is 1.67 (calculated as -log[NaOH]).
The pH of a 0.002M solution of NaOH is around 11.98. This is because NaOH is a strong base that dissociates completely in water to produce hydroxide ions, which lead to the alkaline pH.
-log(10^-4 M NaOH) = 4 14 - 4 = 10 pH NaOH -----------------------
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)
The pH of a 0.1M NaOH solution is around 13. NaOH is a strong base that dissociates completely in water to produce hydroxide ions, leading to a highly alkaline environment with a high pH value.
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.
How to work it out - Calculate the concentration of the solution in terms of molarity. Calculate the hydrogen ion concentration from the fact that multiplying the hydrogen and hydroxide concentrations (in mols per litre) will always give 1x10-14. Take the -log10 of the hydrogen ion concentration.