Can you calculate the ph of 0.25 M NaOH?

Question

and pls..can you calculate this one?

calculate the pH of 0.050M HNO3

Answer 1

You use your calculator:

Shift - Log (10^x) then you punch "-" (negative) then the concentration - [H+(concentration)]

For 0.25 M NaOH punch it like this:

(Shift-Log (becomes a small 10)) 10-0.25

pH of 0.25 M of NaOH = 0.56

0.050 M HNO3

10-0.050

pH of 0.050 M of HNO3 = 0.89

Answer 2

NaOH----->Na+ + OH-

since it is a strong base the [OH-]=0.25 pOH=-log[OH]=-log[0.25]

pH + pOH=14 you can find using these.

HNO3 is also a strong acid so pH=-log[H+]=-log[0.050]

Answer 3

pH = log ( 1/ [ H+] )

pOH = log (1/ [ OH-])    

pH + pOH = 14

for the first case u are calculating OH- concentration, and it's the same concentration because its a strong base

pOH = log (1 / 0.25)  = 0.6

pH = 14- pOH

pH= 13.4

The other case its a strong acid

pH = log ( 1/ 0.05) =  1.3

You have to keep in mind ALWAYS  if u r calculating pH or pOH for strong or weak bases or acids

hope it works :P

Answer 4

pOH = -log[OH-] = -log(0.25) = 0.60

pH + pOH = 14

pH = 14.00 - 0.60 = 13.40 for the 0.25 M NaOH

pH = -log[H+] = -log(0.050) = 1.30 for the 0.05M HNO3

====== Follow up ========

I have no idea what Mystx is doing, and I'm pretty sure he doesn't either.

I thought it might be helpful to add a bit about the reporting pH and significant digits.  The only digits which reflect the precision in a pH measurement are those to the right of the decimal.  Consequently, if the concentration is reported to two significant digits, then the pH must be reported to two decimal places.

The digits to the left of the decimal place don't reflect the precision in the measurement.  They come from the exponent in 10 raised to the whatever power.

That is why the pH of the NaOH is 13.40.  It has two significant digits and those are the .40.  The same is true for the pH of the HNO3.  The significant digits are .30.