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Yes, oxalic acid can be titrated by HCl because oxalic acid is a diprotic acid and can react with HCl in a simple acid-base reaction. The titration involves determining the volume of acid required to neutralize the oxalic acid solution, which can be used to calculate the concentration of oxalic acid.
As many as hydroxide concentration:[OH-] = 10-[14.0-pH] = 10-[14.0-11.2] = 10-2.8 =invlog(-2.8) = 1.6*10-3 mol/LSo, to neutralize you've to add 1.6*10-3 mole strong monoprotic acid (H+) to 1.0 L solution of pH 11.2
The answer is o,13 g KOH.
Strong base is added to neutralize the strong acid (H2SO4).
The primary indicator used in Total Base Number (TBN) testing is the amount of acid (often sulfuric acid) required to neutralize the basic components present in a lubricant sample. This indicates the alkalinity reserve of the lubricant, which can help determine its ability to neutralize acids formed during combustion in an engine.
Well,i guess the same volume as the base.
To determine the volume of potassium hydroxide solution needed to neutralize the hydrochloric acid solution, you can use the formula M1V1 = M2V2. By plugging in the given values, you can calculate the volume of the potassium hydroxide solution required. In this case, the volume of the 0.152 M potassium hydroxide solution needed to neutralize 10.2 ml of the 0.198 M hydrochloric acid solution would be 7.43 ml.
To determine the volume of calcium hydroxide needed to neutralize the nitric acid, you can use the equation n1v1=n2v2, where n is the number of moles and v is the volume. As the concentration and volume are given for both the acid and base, the volumes of both solutions needed to neutralize each other will be equal. Therefore, the volume of 0.0550 M calcium hydroxide required will also be 35.00 mL.
The hypothesis of an acid-base titration is that the volume of the acid solution needed to neutralize a base solution is stoichiometrically equivalent to the volume of the base solution required to neutralize the acid. This forms the basis for determining the unknown concentration of an acid or base by titration.
To determine the volume of NaOH solution needed to neutralize an acid solution, you would need to know the concentration of the acid solution and the volume of the acid solution. Using the equation n1V1 n2V2, where n represents the number of moles and V represents the volume, you can calculate the volume of NaOH solution needed.
To neutralize the acid, we need to use the same number of moles of base. First, calculate the number of moles of HCl using its concentration and volume. Then, use the mole ratio from the balanced equation to find the required volume of NaOH. Convert the volume to mL.
Use this to analyse the unkown and variables:Ca*Va = Cb*Vb in which:C = concentration (mol/L = mmol/mL)V = Volume (L or mL)subscripts: 'a' = acid, 'b' = baseWanna know Vb ? well find the other three variables to calculate: (Ca*Va)/Cb = Vb (with the same unit as used for Va)
Yes, oxalic acid can be titrated by HCl because oxalic acid is a diprotic acid and can react with HCl in a simple acid-base reaction. The titration involves determining the volume of acid required to neutralize the oxalic acid solution, which can be used to calculate the concentration of oxalic acid.
As many as hydroxide concentration:[OH-] = 10-[14.0-pH] = 10-[14.0-11.2] = 10-2.8 =invlog(-2.8) = 1.6*10-3 mol/LSo, to neutralize you've to add 1.6*10-3 mole strong monoprotic acid (H+) to 1.0 L solution of pH 11.2
In an acid-base titration, a known volume of acid or base of unknown concentration is titrated with a standardized solution of base or acid of known concentration, respectively. The setup involves adding an indicator to the solution being titrated, which changes color at the endpoint when the reaction is complete. The volume of the standardized solution required to neutralize the unknown solution is used to calculate its concentration.
The largest volume of 0.100M sodium hydroxide solution would be needed to neutralize a strong acid with a low molarity. This is because a lower molarity acid would require more moles of sodium hydroxide to neutralize it, resulting in a higher volume of the solution being needed.
The answer is o,13 g KOH.