pH=-log[H+]
pH=-log[1.00x10^-5M]
pH=5
When concentration of H+ is 1M, the pH is zero. Every time the concentration is decreased by a magnitude of 10 (i.e. 10^1 ---> 10^-2) the pH goes up 1 value. This is true for the pH of bases as well, but in that case pH=14+log[OH-], which is derived from pOH=-log[OH] and pH+pOH=14. This is true for pure water at 25 degrees Celsius.
[OH-] = 1e-12
pOH = 12
pH = 2
By definition pH = -log [H+] (in water) so the answer is found by calculation of:
pH = -log [1.2 * 10-2] = 1.92
pH = -log[H] = -log(1.0 x 10^-5) = 5
10
31M
4.0%
n=c/v n=3M/.25L n=12 mol m=Mxn m=58.443 g/mol x 12 mol m=701.3 g n= mol c=concentration v=volume m=mass M= molar mass Tylerops: I don't agree with this answer. Molarity is defined as Moles/Liters. In other words Molarity is the concentration of a solution. In the above n= Concentration / Liters. That is equal to saying Moles=(Moles/liters)/ Liters. In the above question the concentration is (3 moles/ liter), or 3M. Plus, how can it be possible to have 12 moles in 250ml when you only have 3 moles in each liter of the original solution? Correct ANSWER: 3.00 M, or 3 moles per (L) "liter" calls for having 3 moles per liter of the solution. The question asks how many moles must be in 250ml of a solution that has 3 moles per Liter. You must ask yourself what percent of 1 Liter is 250mls? Since there are a thousand ml in one liter, (1000ml=1L), then 250ml is exactly 25% of a Liter, or .25L. So, 250ml can only hold 25% of the 3.00 Molarity. Meaning that you multiply 3 x .25 and get .75 moles. 58.443g/molNaCl x .75 moles = FINAL ANSWER 43.83225g NaCl, Sig Fig, 43.83gNaCl
(AuCl has Ksp = 2.0 x 10-13)-1.0 x 10-12
ask your chemistry teacher at 12 PM tomorrow
pH 2
since it is a strong alkali its pH value will be 12-14 Depending on the concentration: 1% solution- ph12 10% solution- ph13 50% solution- ph14
It mainly depends on the concentration and will approx. range from 10 to 12
- log(2.3 X 10 -12 ) = 11.6 pH -----------------very little H + concentration in this solution.
1/1011.27 = 5.370 x 10 -12 ============
It can be either. Two examples would be: Caustic soda (Sodium Hydroxide), which is very alkaline. Caustic potash (Potassium Hydroxide), which is very acidic.
[h+]=1 * 10-2[h+][oh-]=1 * 10-14[oh-]=1 * 10-12
- log(2.7 X 10 -3 M HCl)= 2.568614 - 2.5686= 11.4314===========now,1/10(11.4314)= 3.7 X 10 - 12 M OH -=================
5 * 10**-12 mol 32 * 10**-9 mol Concentration (M) * Volume (L) = mols C1*V1=C2*V2 (5*10**-12)*V1=(32*10**-9)*V2 (5*10**-12)*V1/(32*10**-9)=V2 (5*10**-3)*V1/32=V2 The volume of the 5 picomolar solution that you wish take = V1 The volume of the 32 nanomolar solution that you need to make V1 at 5pM concentration = V2 Take V2, and place into graduated cylinder and fill to V1.
- log(2.3 X 10 -12 ) = 11.6 pH -----------------very little H + concentration in this solution.
A solution with a pH of 12 is very basic (alkaline). The concentration of hydrogen ions (H+) is 10-12 moles per liter and the concentration of hydroxide ions (OH-) is 10-2 moles per liter. Household ammonia has a pH of about 12.
What is the pH of sodium hydroxide? What I determined from a wide range pH paper is that the pH of a .1 M solution of sodium hydroxide was that between 11 and 12.