A 500-ml sample of 0.100 M formate buffer, pH 3.75 is treated with 5 ml of 1.00 m KOH. pH?
HA + OH^- => A^- + H2O 50.....5...........50 45....0...........45 so pH = 3.74 - lg(45/55) = 3.83
A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH. Determine the pH of the solution after the....?
HF + KOH --> KF + H2O 0.1L x 0.2M HF = 0.02moles HF 0.1L x 0.1M KOH = 0.01moles KOH 0.01 moles HF will be neutralized by the KOH leaving 0.01moles HF in a total of 0.2L volume 0.01moles HF / 0.2L = 0.05M HF Ka = 3.5x10^-4 = [H+][F-] / [HF] 3.5x10^-4 = x^2 / 0.05 - x x^2 + 3.5x10^-4x - 1.75x10^-5 = 0 x = [H+] = 0.00401M pH = -log[H+] = 2.4 1) determining just the pH of the HF we would get 2.09 2) adding 1/2 the amount of OH- as requires to neutralize HF leaves 1/2 the moles of HF, which is still acidic. the pH wouldn't get close to 3.46 with only 0.1M H+, even from the weak acid HF.
A 500-mL sample of a 0.100M formate buffer, pH 3.75, is treated with 5 mL of 1.00M KOH. What is the pH?
Well with Potassium Oxygen and Hydrogen the pH would be 4.5.
A 0.1g sample of CaCO3 is treated with 20ml of 0.1N HCl solution and excess of HCl is neutralised with 5ml of 0.1M Ca(OH) 2. What is the percentage of purity of CaCO3 sample?
CaCO3 +2HCl ———-> CaCl2 +CO2 +H2OSo 1 mole of CaCO3 reacts with = 2 moles of HCl (Here for HCl Molarity=Normality)100 gm of CaCO3 reacts = 2 moles of HCl100 gm CaCO3= 1 mole=1000 milli mole0.1 gm CaCO3= 1 milli mole20 ml.of 0.1 N HCl=2 milli mole.So if CaCO3 were 100% pure would have been completely neutralised by 20 ml.of 0.1N HCl.But since it is impure the extra acid present is neutralised by 5 ml.of 0.1 M, Ca(OH)2, which neutralises 10 ml.of 0.1N HCl.Ca(OH)2 +2HCl→2CaCl2 +2H2OSo out of 20 ml of 0.1N HCl, 10 ml is consumed by CaCO3 and the rest unreacted 10 ml is consumed by Ca(OH)2 solution.As I already told if CaCO3 were 100% pure all the 20 ml.of 0.1N HCl, would have been completely consumed by 0.1HCl itself.Since it is impure, it had consumed only 10 ml.of it. So the percentage of purity of CaCO3 is 50%.Otherwise the pure CaCO3 present in the sample amounts to= 0.1 x 10/20 = 0.05gmSo % purity = 0.05 x 100/0.1 = 50.
What is the pH of a buffer if it is made by mixing a 200ml 0.1mol/lit-Nh4Cl solution and a 250ml 0.2mol/lit solution if the kB is 1.8 10-5?
First of all, this isn’t a buffer solution. It is an acid dissociation. The acid being NH4Cl. The solution itself will have the following concentration.(0.1 M NH4Cl) (0.200 L) = 0.02 mol NH4Cl(0.2 M NH4Cl) (0.250 L) = 0.05 mol NH4ClSo, there is a total of 0.07 mol NH4Cl in a total volume of 200 mL + 250 mL = 450 mL so, the concentration of NH4Cl would be:0.07 mol NH4Cl/0.450 L = 0.1556 M NH4ClThe acid dissociation equilibrium reaction is:NH4+ <-> NH3 + H+Whose Ka would be:Ka = Kw/Kb = 1x10^-14/1.8x10^–5 = 5.55x10^-10So, based on the concentration of NH4+ and the Ka for the reaction, the pH would be:x^2/(0.1556-x) = 5.55x10^–10x = 9.3x10^–6 M H+pH = 5.03
Calculating the pH of a buffer solution?
Hi, well this isn't so bad I would be glad to help. First off we need to decide which of these is our acid and which is the base. HCHO2 is formic acid - so it's the acid NaCHO2 is sodium formate. In an aqueous solution it is present as a sodium ion (Na+) and a formate ion (CHO2-) because of the negative charge on the formate it can accept an H+ and is a base. Now since we're working with a buffer we can use the Henderson–Hasselbalch equation: pH = pKa + log (base/acid) Quick note - in this equation you can use either moles of base and acid or concentration of base and acid, but you can only use concentration if the base and acid are contained in the same volume. Since the base and acid are in different volumes here we have to solve for the number of moles of each. To find the moles of base (sodium formate) we do this: 15 mL x (1 L / 1000 mL) x (0.5 mol / 1 L) conversion to L molarity of base When you solve you get : 0.0075 mol sodium formate Do the same for acid (formic acid): 60mL x (1L / 1000 mL) x (0.250 mol / 1 L) And you get: 0.015 mol formic acid Now we have all of the parts of the equation we need except pH (which we're solving for) and pKa. We HAVE to have the pKa and luckily for us it's a physical constant so you can look it up anywhere. Happens to be 3.744. Let's start plugging information into the equation: pH = 3.744 + log (0.0075/0.015) On your calculator make sure you do the log first if you have a scientific calculator or enter it like this 3.744+(log(0.0075/0.15)) if you have a graphing calculator. If you enter it correctly you get pH = 3.44! Hope this helped!
Acetic Acid - Sodium Acetate Buffer?
pKa = - log 1.8 x 10^-5 = 4.7 4.8 = 4.7 + log [acetate]/ [acetic acid] 10^0.1 =1.3 = [acetate] / [acetic acid] [acetate]= 1.3 M [acetic acid]= 1 M M1V1 = M2V2 0.100 V = 1 M x 0.0600 L V =0.600 L
Determining pH?
Moles acid = 0.30 x 0.500 L =0.150 Moles acetate = 0.20 x 0.500 = 0.100 CH3COOH + OH- >> CH3COO- + H2O Moles OH- = 0.020 L x 1.00 = 0.020 moles acid = 0.150 - 0.020 =0.130 moles acetate = 0.100 + 0.020 =0.120 Total volume = 500 + 20 = 520 mL = 0.520 L concentration acid = 0.130 / 0.520 = 0.250 M concentration acetate = 0.120 / 0.520 = 0.231 M pKa = 4.74 pH = 4.74 + log 0.231 / 0.250 = 4.71
500 ml of 0.2M BOH (a weak base) is mixed with 500 ml of 0.1 M HCl and the pH of the resulting solution is 9. What is the pH of a 0.1 M BCl solution?
This uses stoichiometry(i think that is how you spell it; i did not learn chemistry in english so i apologize in advanced for the mispelling in the jargons).The reaction:BOH + HCl → BCl + H20 (the reaction is already balanced)From the pH result we can conclude that the concentrace of H+ from the BCL is 10^-9 which makes the concentrate of the OH- (the base) is 14 - 9 = 5, so [OH-] = 10^-5We are using the Buffer formula for weak bases which is:[OH-] = Kb x [OH-; from BOH] : [H+; its conjugation acid from HCl]*Kb= the constant value of the base BOH since it is not provided, we use the Buffer formula to find ke Kb and then find the concentration to figure out the pH of 0,1M of BClPut all the known data into the Buffer Formula: (as follows)10^-5 = Kb x 0,5 mol : 0,5 molThe Kb is 10^-5*find the mol for each of the compoundsBOH = 0,5 litres x 0,2M = 0,1 molHCl = 0,5 litres x 0,1M = 0,05 molThe BOH will react with the value of 0,05mol because that is the balanced constant; which leaves the BOH to 0,05 mol (0,1–0,05)Now that we have the Kb,We use the formula for determining the value of a weak base which is;[OH-]^2 = Kb x [OH-]*for the [OH-] that is multiplied with the Kb we use the information from the 0,1M BCl, because the konstanta of the reaction is 1 (which means the reaction is balanced) we can use 0,1M as it isThe result is[OH-]^2 = 10^-5 x 0,1M[OH-] = 10^-3 → pOH = 3 → pH = 14–3 = 11The pH of 0,1M BCl is 11(Hope this helps, sorry for any mistakes)