how do i do this problem. it says PH= - log [H+] and [ H+]= 4.2x10^-8
do i just plug it in or wat?
2007-12-10 09:44:10 · 6 answers · asked by NoName 3 in Science & Mathematics ➔ Chemistry
Yup. Just plug it in and work the numbers. Take the base-10 logarithm of 4.2*10^-8 and change the sign to get the pH.
log(4.2*10^-8) = -7.38
pH = -(-7.38) = 7.38
Strictly speaking, there is a problem with defining the pH in this way. The square brackets are used to indicate "the concentration of" something (in this case of the H+ ion), but one can't take the logarithm of a quantity that has units, and concentration most definitely has units. In reality, the pH is defined as the negative logarithm of the "thermodynamic activity" of the hydrogen ion. The activity is a measure of the "effective concentration" of a solute, and takes into account interactions between the molecules of the solute and the solvent, as well as other solutes. Activity is a unitless quantity, and is always expressed relative to a standard state, which for aqueous solutions, is normally taken as 1 molar. With this convention, the activity of an dilute solution (one that behaves "ideally") is numerically equal to the molar concentration, but without the units of concentration.
2007-12-10 09:47:44 · answer #1 · answered by hfshaw 7 · 1⤊ 0⤋
Allow me add a word about pH and significant digits. The only digits that carry any precision in a pH are those to the right of the decimal. For instance, a pH of 7.2 has one significant digit. A pH of 12.34 would have two significant digits. The "7" or the "12" in the examples are not significant. They come from the exponent in the original hydrogen ion concentration. When Trevor writes [H+] = 6.31*10^-8 and [H+] = 1.58*10^-8 for pH values of 7.2 and 7.8, respectively, he is writing them with more precision than is justified. The hydrogen ion concentration for a pH of 7.2 or 7.8 should be written to only one significant digit, as in 6*10^-8 and [H+] = 2*10^-8.
2016-04-08 06:41:53 · answer #2 · answered by Anonymous · 0⤊ 0⤋
pH = - log([H+] = log(1/[H+])
When working with log tables, it is easier to take the log of 1/[H+]. With a calculator just enter the [H+] and take the log, then change its sign.
2007-12-10 10:13:48 · answer #3 · answered by skipper 7 · 0⤊ 0⤋
PH is also = to 1/log[H+] , now plug it in.
2007-12-10 09:56:55 · answer #4 · answered by jacob s 7 · 0⤊ 1⤋
yup and the result will be 7.38 which means it will be basic.
2007-12-10 09:49:49 · answer #5 · answered by greeK_god_HirYu 3 · 0⤊ 0⤋
-log 0.05
2014-03-19 02:29:20 · answer #6 · answered by allwar a 1 · 0⤊ 0⤋