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Preparation of phosphate buffer pH 3?
Volumetric flask= 200 ml (100 ml K2HPO4 0.1M)+ (44.6 ml HCl 0.1M) added in flask then added deionized or distilled water until mark.
Calculate the pH of a solution prepared by diluting 3.0 mL of 2.5 M HCl to a final volume of 100 mL with H2O?
Because HCl is a strong acid, it dissociates completely to H + Cl. Therefore, 3.0 mL x 2.5 M HCl = 7.5 meq of H+ . In 100 mL of solution, this is 0.075 M H+. pH= - log [H] = - log (0.075) = - (-1.1) = 1.1 (two significant figures)
How would you prepare 1600 ml of a pH 1.50 solution using concentrated 12M HCl?
Start with the pH. pH is -log(H)=1.5 Solve for H (this is your concentration of hydrogen). This comes out to be .0316=[H]. This means there are .0316 mols of H in 1 liter of solvent. So now that we know this, we can use this equation MoVo=MfVf (o is initial, f is final, M is molar, V is volume) So plug in the numbers (.0316)(1.6)=(12)(Vf) Vf = .001813 L or 1.813 mL. So you'd take 1.813 mL of 12 M HCl and dilute it to 1600 mL This makes sense because 12M HCl has a very LOW pH, so diluting with such amount of solvent is necessary. Now that I have more time, I'll tell you how you want to approach this. The thing is all about the equation MoVo=MfVf. This equation is used for diluting. Now we have two of these values, we just need the other M to solve for volume of 12 M HCl. So how does pH relate to this problem (namely concentration)? We pH is a measure of concentration. That is -log[H]. So We must solve for [H]. Upon doing so, we then consider this: What happens when HCl dissociates? Well here it is HCl ====> H + Cl So for every mol of HCl, we get a mol of H. Therefore, [HCl]=[H] So there you have it!!
What is the pH of a 0.280 molar HCl solution?
pH = -log(0.280) = 0.553
What is the pH of a 4.3 x 10-2 M HCl is the solution?
p(x)=-log([x])So... pH=-log([H+])pH = 1.4
What is the pH of 10.0 mL of 0.0020 M HCl?
2.70
Preparation of phosphate buffer pH 3?
Volumetric flask= 200 ml (100 ml K2HPO4 0.1M)+ (44.6 ml HCl 0.1M) added in flask then added deionized or distilled water until mark.
Consider the titration of 30.0 ml of 0.030 M NH3 with 0.025 M HCl the equivalence point is reached when 36 ml of HCl titrant is added what is the pH at the equivalence point?
pH = 5.56 at 36.0 ml(equivalence point)
How do you prepare 0.5M Tris-HCL pH 7.2?
for 1L 121.1 gr tris base 80 ml HcL 920 ml Distilled Water
What is the pH of Na2HPO4?
For example, to obtain a solution with the pH=7,00 mix: 756 mL 0,1 M solution of Na2HPO4 with 244 mL of 0,1 M HCl solution.
Calculate the pH of a solution prepared by diluting 3.0 mL of 2.5 M HCl to a final volume of 100 mL with H2O?
Because HCl is a strong acid, it dissociates completely to H + Cl. Therefore, 3.0 mL x 2.5 M HCl = 7.5 meq of H+ . In 100 mL of solution, this is 0.075 M H+. pH= - log [H] = - log (0.075) = - (-1.1) = 1.1 (two significant figures)
What is the pH of a 0.100 molar HCl solution?
The answer depends on several unspecified variables, most importantly the final molarity of the solution, which depends on the final volume. You can calculate the value yourself using the formula: pH = -log[H+] where [H+] is the final concentration of H+ ions in solution. For HCl, [H+] is equal to molarity. So, for example, if you add 50.0 ml of 1.0M HCl to 950 ml of deionized water, your final concentation is: (50.0 ml/1000 ml) * (1.0M) = 0.05M Therefore: pH = -log[0.05] = 1.3
What is the pH of .25M HCl?
- log(0.25 M HCl) = 0.6 pH ------------
What is the pH of a 1.0x10-4m hcl solution?
- log(0.00450 M HCl)= 2.3 pH=======
What is the pH of 260M HCl?
.260 M of HCL, not 260 More than likely correct, but, - log(0.260 M HCl) = 0.6 pH ----------- ( pH can be below 1 )
How do you make a sodium phosphate buffer solution of pH 6.8?
For pH = 7,0: 756 mL disodium hydrogen phosphate (Na2HPO4) solution 0,1 M (14,2 g/L) + 244 mL hydrochloric acid (HCl) solution 0,1 M
Is pH 2.0 HCl the same with 0.1 M HCl?
pH=-lg[H+][H+]=10-pHWith pH=2.0:Corrected:[H+]= 10-pH = 10-2.0 = 0.010 M HCL
