14.
Calculus:
pOH = -log(conentration of OH-)= -log(1.0)= 0.0
pH = 14-pOH = 14 -0.0 = 14.0 (at 250C)
-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.
THE PH VALUE ACIDIC SOLUTION VARIOUS FROM 0-6.9, WHILE THE BASIC SOLUTION VARIOUS FROM 7.1-1.4. THUS ,OUT OF HCL AND NaOH WILL HIGHER PH VALUE
The pH of 0.1 M NaOH is approximately 13, as NaOH is a strong base that fully dissociates in water to produce hydroxide ions. The pH of 0.05 M NaOH would also be approximately 13, as the concentration of the base does not impact the basicity significantly when dealing with strong bases.
The pH of a 0.1 M NaOH solution is approximately 13. NaOH is a strong base that dissociates completely in water to produce hydroxide ions, which results in a highly alkaline solution with a pH greater than 7.
-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.
THE PH VALUE ACIDIC SOLUTION VARIOUS FROM 0-6.9, WHILE THE BASIC SOLUTION VARIOUS FROM 7.1-1.4. THUS ,OUT OF HCL AND NaOH WILL HIGHER PH VALUE
The pH of a 1 millimolar NaOH solution is approximately 11. The concentration of a 1 millimolar solution is 0.001 mol/L, and NaOH is a strong base that completely dissociates in water to produce hydroxide ions, leading to a basic pH.
The pH of 0.1 M NaOH is approximately 13, as NaOH is a strong base that fully dissociates in water to produce hydroxide ions. The pH of 0.05 M NaOH would also be approximately 13, as the concentration of the base does not impact the basicity significantly when dealing with strong bases.
The pH of a 0.1 M NaOH solution is approximately 13. NaOH is a strong base that dissociates completely in water to produce hydroxide ions, which results in a highly alkaline solution with a pH greater than 7.
pH = -log10[H] so 1.96 x 10-2 = -1.71 so 1.71. pH + pOH = 14 14 - 1.71 = 12.29
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.
The pH of a 0.33 M solution of NaOH is approximately 13.5. This is because NaOH is a strong base that dissociates completely in water to give hydroxide ions, which results in a highly alkaline solution.
Use the working definition of pH used in General Chemistry classes: pH = -log([H+]) and the equilibrium constant for ionization of water: [H+][OH-]=10-14 (Here [] denotes concentration in Molarity) For moderate concentrations of NaOH (like 10-4-ish M and up, we can neglect the [OH-] from the actual ionization of water (since 10-7 is the maximum this concentration could be, when the NaOH concentration is 0, and even this is much less than the concentration of NaOH). Then we can say: 10-14=[H+][OH-]=[H+][NaOH] and then pH=-log[H+]=-log(10-14/[NaOH]) Just as an example, a 0.5 M solution has a pH of approximately -log(10-14/0.5) which is about 13
The pH of 1 M NaOH is around 14. Sodium hydroxide is a strong base that completely dissociates in water to produce hydroxide ions, which results in a highly alkaline solution with a pH at the upper limits of the pH scale.
To find the pH of a 0.6 M NaOH solution, first, note that NaOH is a strong base that dissociates completely in water. The concentration of hydroxide ions (OH⁻) will also be 0.6 M. The pOH can be calculated as -log(0.6), which is approximately 0.22. Since pH + pOH = 14, the pH of the solution is about 13.78.