how can claculate this?
what is the formula?
which book is better idea for this?
2006-07-23 03:00:09 · 8 answers · asked by panneer s 1 in Science & Mathematics ➔ Chemistry
pH=-log[H+]
at room temperature the equilibrium constant for ionization of water is Kw=[H+][OH-]=10^-14
Therefore -log([H+][OH-])=-log(10^-14) =>
(-log[H+])+(-log[OH-])=14
and pH+pOH=14
5N NaOH means [OH-]=5 M
That would mean pH=14-pOH=
=14-(-log5)=14+0.69=14.69
This is an unrealistic value (the range is 0-14 at room temperature) because pH is a measure of dilute aqueous solutions (N<1). You are not meant to use pH for N>1...
Any chemistry text book that has an intro on acids,bases and pH is good to help you with this.
2006-07-23 03:14:07 · answer #1 · answered by bellerophon 6 · 2⤊ 1⤋
The pH will simply be >14 since the pH scale is not defined for solutions that concentrated. You have to go to a different scale when you get away from dilute solutions.
2006-07-23 05:19:59 · answer #2 · answered by Peter Boiter Woods 7 · 0⤊ 0⤋
"common" is perplexing, yet that doesn't count right here, with the aid of fact acetic acid provides in basic terms one H+, and NaOH provides in basic terms one OH-, in line with formulation unit, so normality and molarity is the comparable element in those circumstances. the two calculations are incorrect. 0.1M stable acid provides pH a million, yet acetic acid is a vulnerable acid, and the pH will certainly be around 2.9. (you will locate out the thank you to artwork that out in case you do vulnerable acid equilibria) 0.a million NaOH, [OH-] = 10^-a million so [H+] = 10^-13 so pH is 13.
2016-10-08 05:45:24 · answer #3 · answered by ? 4 · 0⤊ 0⤋
You can't have a solution of NaOH with a pH of 5. To go from pure water, pH 7, to pH 5, you have to increase the number of solvated protons. NaOH has none to give.
2006-07-23 08:23:16 · answer #4 · answered by rb42redsuns 6 · 0⤊ 0⤋
pH + pOH = 14 for water.
5N NaOH will completely dissociate and produce a pOH of -0.69897. This is because pOH = -log[OH] = -log(5)
rearranging the first equation we get
pH = 14 - pOH
pH = 14 - (-0.69897)
pH = 14.699
2006-07-26 09:16:59 · answer #5 · answered by Anonymous · 0⤊ 0⤋
we know that pH+pOH = pKw
pKw=-logKw
at room temperature i.e at 25oC Kw=H+ + OH-
Normality of NaOH is 5N
As acidity of NaOH is 1 molarity of NaOH is 5M
we know that normality of a monovalent base=molarity=OH- ion concentration.
[OH-]=5
pOH=-log [OH-]
= -log(5)
= -0.6989
pH + pOH = pKw
at room temperature pKw=14
pH=14-pOH
=14-(-0.6989)
=14+0.6989
=14.6989
The book : you can refer analytical inorganic chemistry
or indian pharmacopio
or p.c kamboj
2006-07-23 05:10:12 · answer #6 · answered by anjs 2 · 0⤊ 0⤋
read ur book carefully.10+2chemistry book can help u.
2006-07-23 03:04:41 · answer #7 · answered by Sayom 2 · 0⤊ 0⤋
pH+poH=1
pH=1-poH
2006-07-23 05:45:31 · answer #8 · answered by Anonymous · 0⤊ 0⤋