pH= -log[H+] (read as the concentration of hydrogen ions). Thus to find [H+], you need to determine the antilog of the reciprical of the pH. For example, if the pH is 3.76, then you need to find the antilog of -3.76 (because log[H+]=-3.76). The concentration of H+= approximately 1.74x10-4.
As for OH-, we first need to establish that the negative log of a quantity is labeled p, so the concentration of OH- is expressed as pOH-, and pH+pOH=14. Here are some helpful examples: If the pH is 3, the pOH- is 11. If the pH is 5, then the pOH- is 9. So basically what you do is subtract the pH from 14 to find the pOH-. Such as in an example where the pH is 4.5. You would have 14-4.5, which equals 9.5.
If your given pH or pOH, you can also find [H+] or [OH-] use antilog
The pH tells you the concentration of H+ ions in the solution according to this formula pH = -log [H+] (where the square brackets mean "the concentration of" whatever is inside the brackets) So, if you have the pH, you can find the concentration of H+ from this: [H+] = 10-pH If the pH is 5.00, then 10-5 = 1 x 10-5 M = 0.00001 moles per liter But that's [H+], not the concentration of [OH-]! But those two are related like this: [H+] * [OH-] = 10-14. So to find [OH-], we use: [OH-] = 10-14 / [H+] In this case, [OH-] = 1 x 10-9 M
Depends on the pH, at low pH (below pH 7), the H+ ion concentration is greater, and high pH the OH- ion concentration is greater
pH= -log[H+] pH + pOH = 14 pOH = 14 - pH pOH= -log[OH], so the antilog of -pOH will give you the OH concentration.
pH = - log([H+]) , pOH = - log([OH-] , pH + pOH = 14 [X] = concentration of X
There is no way to know the pH of sodium bicarbonate unless the concentration of [H+] or [OH-] is known. If the concentration is known, pH can be calculated as the -log[H+], or 14-(-log[OH-]).
If your given pH or pOH, you can also find [H+] or [OH-] use antilog
The pH tells you the concentration of H+ ions in the solution according to this formula pH = -log [H+] (where the square brackets mean "the concentration of" whatever is inside the brackets) So, if you have the pH, you can find the concentration of H+ from this: [H+] = 10-pH If the pH is 5.00, then 10-5 = 1 x 10-5 M = 0.00001 moles per liter But that's [H+], not the concentration of [OH-]! But those two are related like this: [H+] * [OH-] = 10-14. So to find [OH-], we use: [OH-] = 10-14 / [H+] In this case, [OH-] = 1 x 10-9 M
ph 3= [H+] = 1 x10^-3 ph 10 = pOH = 4 = [OH-] = 1 x10^-4 2 x (1 x10^-3) = 2 x10^-3 mmol H+ ions = 0.002 mmole 3 x (1 x 10^-4) = 3 x10^-4 mmol OH- ions = 0.0003 mmole H+ > OH- so subtract to get remaining OH- 0.002 - 0.0003 = 0.0017 mmole H+ find concentration [H+] = mmole H+ / ml solution = 0.0017 / (3+2) = 0.00034 M H+ now find pH pH = - log [H+] = - log(0.00034) use a calculator to find the answer.
Depends on the pH, at low pH (below pH 7), the H+ ion concentration is greater, and high pH the OH- ion concentration is greater
pH= -log[H+] pH + pOH = 14 pOH = 14 - pH pOH= -log[OH], so the antilog of -pOH will give you the OH concentration.
pH = - log([H+]) , pOH = - log([OH-] , pH + pOH = 14 [X] = concentration of X
pH depends on ions H+ or OH-.
The concentration of H+ or OH-.
Ions H+ and OH-.
Its hydrogen ion concentration determines whether a solution is to be acidic (pH below 7) or basic (pH higher than 7.0) .Acidic: [H+] > [OH-] , or pH < 7.0 in water at 25 oCBasic : [H+] < [OH-] , or pH > 7.0 in water at 25 oCIn between is neutral: [H+] = [OH-] = 1.0*10-7, or pH = 7.0 in water at 25 oCNote:Remember that (always, by definition) pH = - log[H+] and[H+] = 10-pHand [H+] * [OH-] = 10-14 (or pH + pOH = 14.0) in water at 25oC
pH + pOH = 14 pH is a measure of the hydrogen ion concentration, [H+] pOH is a measure of the hydroxide ion concentration, [OH-] pH = -log10[H+] pOH = -log10[OH-]