Suppose you have 49.5 mL of a solution of H2SO4. You titrate it with a 0.10 M solution of RbOH and use 15.4 mL to reach the endpoint...?
H2SO4 + 2RbOH -------> Rb2SO4 + 2H2O No. of moles of H2SO4 = 1/2 x No. of moles of RbOH = 1/2 x 0.10M x 0.0154L = 7.70 x 10^-4 mole But No. of mole of H2SO4 = 7.70 x 10^-4 = molarity x volume = M x 0.0495L M = (7.70 x 10^-4Mol)/0.0495L = 0.016 mol/L [H2SO4] = 0.016M hence [H+] = 2 x 0.016 = 0.032mol/L pH = - log[OH-] = 1.50
Suppose you have 49.9 mL of a solution of HF. You titrate it with 0.10 M solution of Ba(OH)2 and use 15.5 mL?
Equation: 2HF(aq) + Ba(OH)2 aq) → BaF2(aq) + 2H2O(l) 2mol HF react with 1 mol Ba(OH)2 Mol Ba(OH)2 in 15.5mL of 0.10M solution = 15.5/1000*0.1 = 0.00155 mol This will react with 0.00155 *2 = 0.0031 mol HF Therefore 49.9mL of HF solution contains 0.0031 mol HF 1000 mL HF solution will contain: 1000/49.9*0.0031 = 0.062mol per 1.0L solution Molarity of HF solution = 0.062M In order to calculate the pH of the original HF solution you must calculate [H+] HF is a weak acid and it dissociates to a limited extent. You use the Ka equation to calculate [H+] Ka for HF = 7.2*10^-4 - you must be given this or you look it up in a table - which must be supplied . You should not be required to memorise Ka values of weak acids. Equation: Ka = [H+]² / [HF] ( because dissociation is small you can use 0.062M for [HF]) 7.2*10^-4 = [H+]² / 0.062 [H+]² = (7.2*10^-4)* 0.062 [H+]² = 4.464*10^-5 [H+] = √(4.464*10^-5) [H+] = 6.68*10^-3M pH = -log [H+] pH = -log (6.68*10^-3) pH = 2.18 Note: I have calculated [H+] using the "approximation method" . If you find that your teacher requires the longer quadratic method - email me and I will do this for you.
Suppose you have 49.5 mL of a solution of HNO3. You titrate it with a 0.10 M solution of KOH and use 15.3 mL to reach the endpoint.?
1) What is the concentration of HNO3? 2) What was the pH of the original HNO3 solution? PLEASE SHOW WORK because I want to understand how you came about the answer, for I am confused where to start with the problem and everything
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) .
What volume of 0.0293 M Ba(OH)2 is required to neutralize 25.00 mL of 0.200 M HNO3?
Absolutely, with no offending, the first person's both ways are wrong. There is no way you can use the formula of C1V1=C2v2 with titration process(neautralization) because if you use it, your assuming that both solutions has the same volume and same concentration which is wrong! you only use this equation during dilution of ONE solution, the volume and concentration of it are equal because moles never change in dilution Solution: write the equation first (always write the equation in any chemistry problem to make everything clear). In fact, especially in this problem, you must write down the equation Ba(OH)2(aq)+2HNO3(aq)--->Ba(NO3)2(aq)+... Products in this equation are not necessary in this problem. First you must get the moles of Ba(OH)2 in order to get the volume. To know the moles of Ba(OH)2, the moles of HNO3 first by multiplying volume in L by molarity as follows: 0.025*0.200=0.005 moles Notice that there is A "2" in front of HNO3 in the equation but there is A "1" in front of Ba(OH)2. This simply means that the moles of HNO3 will be DOUBLE the moles of Ba(OH)2 so to get the moles of Ba(OH)2 we will divide 0.005/2=0.0025 moles Now you know the volume of Ba(OH)2... its easy to get its volume Moles=volume(L)*concentration(M) 0.0025=x*0.0293.... doing the simple mass to get the X, which is the volume, 0.0025/0.0293=0.085 L or 85 ML I'm 100% sure of my answer...(I always get A's in tests in chemistry because its my favourite subject) I wrote too much but I just wanted to make sure you understand I hope that helped and good luck!
What is the pH of a 0.001 M solution of rubidium hydroxide, RbOH, a strong base?
THANK YOU to Jacob Bumpus and to skipper!!! I can't get it to show the "pick as best answer" button.... And sorry, but I have three more!! My computer wont load the full page, so I can't ask them as separate questions... What is the pH of a solution that contains 0.10 mol of HCl in a volume of 100.0 L? The concentration of hydronium ions in a certain acid solution is 100 times the concentration of hydronium ions in a second acid solution. If the second solution has a pH of 5, what is the pH of the first solution? What is the molar concentration of hydroxide ions in a solution with a pH of 6?