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What is the pOH of the solution in the question I asked before this one?
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Look at the KOH problem.
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Similar Questions
Complete this table of values for four aqueous solutions at 25 °C.
Solution A: [H+]=0.0093M [OH-]=? pH=? pOH=? Solution B:
Top answer:
All of these follow two/three formulas. (H+)(OH^-) = Kw = 1E-14 pH = -log(H^+) pOH = -log(OH^-) pH +
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Complete this table of values for four aqueous solutions at 25 °C.
Solution A: [H+]=0.0093M [OH-]=? pH=? pOH=? Solution B:
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See above.
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And what is your problem? a. pOH = -log(OH^-). Substitute and solve. b. Use pH + pOH = pKw = 14 and
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Sometimes the basicity of a solution is reported using pOH (pOH = -log[OH-]). What is the
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If you know H^+ or OH^- the other can be calculated from (H^+)(OH^-) = Kw = 1E-14 If you want pH pH
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you're :)
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these values are negative logarithms of the concentrations so a bigger number means a lower
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The pH=8.60 for a 0.016 molar NTD (toluidine) solution. Find Kb of NTD.
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I don't think you did anything wrong. 9.9E-10 rounded is 10E-10 = 1E-9 fpr Kb. I looked up with a
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What are the concentrations of [H+] and [OH-] in each of the following. Determine the pH and pOH for each solution.
A. 1.25 M HCl
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1.25 M HCl = 1.25 M H^+; therefore, Kw = 1 x 10^-14 = (H^+)(OH^-). You know Kw and (H^+). Calculate
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Calculate the mass of strontium hydroxide required to make a 450-mL solution with a pOH of 4.00.
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The pOH is 10. To convert pOH = 10 to OH^-, it is pOH = -log(OH^-) 10 = -log(OH^-) -10 = log(OH^-)
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hey guys I posted this awhile ago im not sure if anyone has read it or whatever but here it is again find the PH of a .00580 M
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To find the pH of a solution of KOH, you can follow these steps: Step 1: Calculate pOH pOH is
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