Concentration of aq. NaOH = 0.25 M
NaOH is strong base and is mono basic in nature. Therefore,
[NaOH] = [OH-] = 0.25 M
Calculate the pOH of the solution
pOH = -log[OH-]
pOH = -log(0.25) = 0.602
Now calculate the pH of soltuionusing following formula
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 0.602 = 13.4
pH of 0.25M solution is 13.4.
3. Determine the pH for a 0.25 M NaOH(aq) solution. Show all calculations in order to...
3. Determine the pH for the two following solutions. Show all calculations in order to receive full credit. a. a 0.45 M CsHsN (aq) solution where Kb = 1.7x10° for CsHsN. b. a 0.35 M HCN (aq) solution where K = 4.9x10-10 for HCN.
A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 15.0 mL of 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titration curve the mixture produced is in, and the pH of the mixture. Assume that the volumes of the solutions are additive. 1)After adding the HCl solution, the mixture is [select one](before, After, at) the equivalence point on the titration curve. 2)The pH of the solution after adding HCl is [select one](7.00,1.40,11.00,12.60).
what is the pH of a solution that is 0.001 M in NaOH? ( show all work)
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