Calculate the [OH-] of each aqueous solution with the following [H3O+] at 25 degrees celcuis?
Just use this equation [H][OH] = 1.0 * 10^-14 Plug in the [H] value for each and solve for [OH] a. [1.0 X 10-8 M][OH] = 1.0 * 10^-14 [OH] = 1.0 * 10^-6 M b. [2.0 X 10-4 M][OH] = 1.0 * 10^-14 [OH] = 5.0 * 10^-11 M c. [5.0 X 10-4 M][OH] = 1.0 * 10^-14 [OH] = 2.0 * 10^-11 M d. [4.8 X 10-12 M][OH] = 1.0 * 10^-14 [OH] = 2.08 * 10^-3 M
Calculate the [OH−] of each aqueous solution with the following [H3O+]. coffee, 1.2×10−5M?
Kw = [H3O+] [OH-] = 1.0x10^-14 [OH-] = 1.0x10^-14 / 1.2x10^-5 = 8.3 x 10^-10 M
Calculate the [OH-] of each aqueous solution with the following [H3O+] .?
This equation: [H+] [OH-] = 1.00 x 10^-14 governs the relationship between hydrogen ion and hydroxide ion in aqueous solution. We simply substitute into the equation and solve: [1.2 x 10^-12] [OH-] = 1.000 x 10^-14 [OH-] = 0.083 M The other three are done in the exact same manner. This equation will also be helpful to you: pH + pOH = 14.000 I simply applied a negative logarithm to each of the three components of the first equation I cited. My acid base stuff: http://www.chemteam.info/AcidBase/AcidBa...
Calculate the [OH-] of each aqueous solution with the following [H3O+]?
NaHCO3 is the formula of baking soda NaHCO3(aq) -----> H+(aq) + (NaCO3)-(aq) H3O+ can also be written as H+. From the dissociation equation of baking soda, 1 mole of NaHCO3 will produce 1 mole of H+. Hence1.2x10^-8 M of NaHCO3 will give you 1.2x10^-8 M of H+ [H+].[OH-] = 1.00x10^-14 [OH-] = 1.00x10^-14 / 1.2x10^-8 = 8.33x10+-7 mol/L
Calculate the [OH-] of each aqueous solution with the following [H3O+]?
Hi Vio, [H3O(+)] x [HO(-)] = Kw = 10^-14 . . . then : [HO(-)] = Kw / [H3O(+)] [HO(-)] = 10^-14 / [H3O(+)] 1/ [HO(-)] = 10^-9 mol/L 2/ [HO(-)] = 8,3•10^-7 mol/L 3/ [HO(-)] = 2,0•10^-5 mol/L 4/ [HO(-)] = 4,2•10^-13 mol/L
Calculate the [H3O+] of each aqueous solution with the following [OH−].??
This is easy ... [H+][OH-] = 1X10^-14 (in all water solutions) so [H+] = (1X10^-14) / [OH-] dishwashing detergent, 1.3×10−3M [H+] = (1X10^-14) / (1.3X10^-3) = 7.69X10^-12 M milk of magnesia, 1.4×10−5M [H+] = 7.14X10^-10 M aspirin, 1.9×10−11M [H+] = 5.26X10^-4 M
Calculate the OH- of each aqueous solution with the following H3O+.?
1. baking soda 1.0*10^-8 2. orange juice 2.1*10^-4 3.milk 4.9*10^-7 4.bleach 4.9*10^-12 I don't have a calculator with the log button, any help is appreciated. I have an 89 and really need these last hmwk pts to push me over the edge... thanks so much!
Calculate the [H3O+] of each aqueous solution with the following [OH-] at 25*C?
It puzzles me why schools recommend that TI30x First of all, the correct formula is [H3O+] * [OH-] = 1 * 10^-14 Always start by putting in the one you know. Problem A ======== 1 EE 14 +/- divide key (can't make that) 1 EE 2 +/- = You should get 1 EE -12 Problem B ======== 1 EE 14 +/- divide by 1.8 EE 11 +/- = You should get 0.000555 or 5.555 EE - 4 Note: === I don't actually own a TI30X but the keys pretty much operate the way I've set this out. The only question is where to put the +/- You might have to enter it as 1 EE +/- 14 divide by 1 EE +/- 2 =
Calculate the [H3O+] of each aqueous solution with the following [OH−].?
Part A dishwashing detergent, 1.7×10−3M Part B milk of magnesia, 1.0×10−5M Part C aspirin, 1.2×10−11M Part D seawater, 2.8×10−6M Please explain how to work. Thanks!