2.70
To prepare 0.1M Tris-HCl, dissolve 1.21 g of Tris (molecular weight 121.14 g/mol) in distilled water to make 100 mL of solution. Adjust the pH to 7.4 using HCl.
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
To prepare 0.5M Tris-HCl pH 7.2, first calculate the amount of Tris base needed using the molecular weight of Tris (121.14 g/mol). Then dissolve the calculated amount of Tris base in water, adjust the pH to 7.2 using HCl while monitoring with a pH meter, and make up the final volume.
The pH of a 6M HCl solution is 0.
The solubility of Ambroxol HCl can vary depending on the specific conditions. Generally, Ambroxol HCl is more soluble in acidic pH than neutral pH. At pH 6.8, the solubility of Ambroxol HCl may be lower compared to acidic pH. Further experimental testing may be needed to determine the specific solubility at pH 6.8.
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)
To prepare 0.1M Tris-HCl, dissolve 1.21 g of Tris (molecular weight 121.14 g/mol) in distilled water to make 100 mL of solution. Adjust the pH to 7.4 using HCl.
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
To prepare 0.5M Tris-HCl pH 7.2, first calculate the amount of Tris base needed using the molecular weight of Tris (121.14 g/mol). Then dissolve the calculated amount of Tris base in water, adjust the pH to 7.2 using HCl while monitoring with a pH meter, and make up the final volume.
The pH of 0.002 M HCl is approximately 2.3 assuming complete dissociation of HCl into H+ and Cl- ions. The calculation involves taking the negative base 10 logarithm of the concentration of H+ ions which is given by the molarity of the acid solution.
The pH of a 6M HCl solution is 0.
I think it'd be pH 7. Same amount of both, providing they are the same molarity!
The solubility of Ambroxol HCl can vary depending on the specific conditions. Generally, Ambroxol HCl is more soluble in acidic pH than neutral pH. At pH 6.8, the solubility of Ambroxol HCl may be lower compared to acidic pH. Further experimental testing may be needed to determine the specific solubility at pH 6.8.
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.
The answer depends on the concentration in the original HCl-sol'n.Let this be Co (mol HCl/Litre). According to dilution 'law': Co.Vo=Cdiluted.VdilutedCdiluted = Co*10.0 ml / 50 mL = 0.20*Co mmol/ml = CdilutedHCl is a monoprotic strong acid, meaning [HCl]:[H3O+] = 1:1H3O+ = concentration HCl solution = Cdiluted = 0.20*Co mmol/ml= 0.20*Comol/LpH = -log10[H3O+] = -log10[0.20*Co] = 0.70 - log10[Co]
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!!
- log(0.00450 M HCl)= 2.3 pH=======