Calculate the pH of 5.00 mL of 0.25 M perchloric acid that has been diluted to 50.0 mL?
I can't figure out what I'm doing wrong -- I can't figure out how to do this. Your help would be much appreciated. Thank you! Calculate the pH of 5.00 mL of 0.25 M perchloric acid that has been diluted to 50.0 mL.
Calculate the pH of a 0.500 M solution of KCN. Ka for HCN is 5.8 x 10^-10?
KCN is a salt of weak acid(HCN) and strong base(KOH). Use the relation for hydrolysis of salt of weak acid and strong base.pH=7+0.5(pKa+logC)Put the values and get answer.
Calculate the pH of the following solutions... ?
1: -log(0.055) = 2: -log(2.8) = 3: 14 - log(3.6) = 4: 14 - log(0.05) = Commentary: the p in pH or pOH is a math operator that means "take the negative log of" pH + pOH = 14. Since #3 and #4 give you [OH-] use pH = 14 - log[OH] In #4 [OH] = 2 x [Ba(OH)2] = 2 x 0.025 = 0.050
Calculate the pH and pOH of an aqueous solution that is 0.035M in HCl and 0.095M in HBr?
Both HCl and HBr are strong acids total concentration of H+ = 0.035 + 0.095 = 0.13 M pH = - log 0.13 = 0.89 ( at two significant figures) pOH = 14 - pH = 14 - 0.89=13
What is the concentration of H+ in 0.031 M HCLO4. what is the pH of the solution. What is the oH- concentratio?
Perchloric acid is a stronger acid than sulphuric and nitric acids. It dissociates completely in solution 1)If [HClO4] = 0.031M Then [H+] = 0.031M 2) pH of solution = pH = -log [H+[ pH = - log 0.031 pH = 1.51 3) Equation [H+] [OH-] = 10^-14 ]OH-] = 10^-14 / [H+] [OH-] = 10^-14 / 0.031 [OH-] = 3.22*10^-13M 4) If you double the [H+] you decrease the pH by 0.30 Mathematically - [H+] = 0.031M *2 = 0.062M pH = -log 0.062 pH = 1.21 Original pH = 1.51 - new pH = 1.21 - change in pH = 0.30
What is the pH of .025 M HClO4-?
The pH is the negative of the log of the hydrogen ion concentration. HClO4 is a strong acid and it is 100% ionized. Therefore the hydrogen ion concentration (or hydronium, if you prefer) is 0.025. pH = -log(.025) = 1.6
What is the pH of a solution by mixing 50 ml of 0.4N HCL and 50 ml 0.2 N NaOH?
Mole = concentration x volumemole of NaOH = 50ml x 0.2 = 10.0mole of HCl = 50ml x 0.4= 20.010 mole of NaOH exactly neutralized by 10 mole of HCl.NaOH + HCl = NaCl + H2Othat is 20 -10 = 10 HCl moleCon of H+ = no of mole of H+ / total volume of solution (50+50 ml = 100 ml)= 10/100 = 0.1pH = log ( H+)= log (0.1) = 1s0 the pH of solution will be one(1) .
Find the pH of each of the following solutions of mixtures of acids?
Second problem: Here is a link to consider: http://www.molecularsoft.com/help/acid_a... The Ka of HF is 6.3 x 10^-4 and that of phenol is 1.00 x 10^-10 In essence, since HF is a much stronger acid, the phenol remains undissociated in solution and contributes no significant hydrogen ion to the solution. So, treat the solution as only HF and perform the calculation. However, a potential wrench in the gears is the high concentration of phenol compared to that of the HF. I don't know enough about this topic (solutions of two weak acids) to hazard a guess as to what happens. First problem: this is a solution of two strong acids, both of which ionize 100% in solution. 1) Determine moles of H+ from each acid: (0.00015 mol/L) (1.00 L) = 0.00015 mol (0.020 mol/L) (1.00 L) = 0.020 mol 2) Sum then up: 0.00015 mol + 0.020 mol = 0.02015 mol 3) Calculate new molarity: 0.02015 mol / 2.00 L = 0.010075 M 4) pH pH = - log 0.010075 = 1.996755 Perhaps, two sig figs seems reasonable ---> 2.00 If you wanted three sig figs ---> 2.000 Third provlem would probably be addressed like problem #2. I did not look up the Ka values. "I don't normally ask this kind of thing here" Ask away! I like ones that cause my brain to work! I'm at an age where I need to be cognizant of the possible advent of dementia (I kid you not!). This type of thinking is helping me in that battle.