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To prepare 2 mol dm-3 HCl solution for titration, you will need to dilute a concentrated HCl solution of known concentration. Calculate the amount of concentrated HCl needed using the formula: C1V1 = C2V2, where C1 = concentration of concentrated HCl, V1 = volume of concentrated HCl needed, C2 = desired concentration (2 mol dm-3), and V2 = final volume of HCl solution. Mix the calculated amount of concentrated HCl with water to reach the final volume. Remember to wear appropriate personal protective equipment and handle concentrated acids with caution.
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23.74cm3 0.0147 mol dm-3NaOHare required to neutralize 25.00 cm3 of HCl calculate concentration of HCl acid in moldm-3and g dm-3?
HCl + NaOH ― NaCl + H2O nNaOH = c*v = 0.01470 * 23.74*10-3L = 3.49*10-4 mol 1 mol HCl reacts with 1 mol NaOH therefore nHCl = 3.49*10-4mol CHCL= nNaOH /VNaOH = 3.49*10-4 / 25.00*10-3 = 13.96 * 10-3 mol.dm-3
What volume of the 0.200mole dm-3hcl is required to neutralize 25.0cm3 of 0.200 moldm-3 BaOH2?
To neutralize the Ba(OH)2, we need an equal number of moles of HCl. The molarity of Ba(OH)2 is 0.200 mol/dm^3, so the number of moles present in 25.0 cm^3 is (0.200 mol/dm^3) x (25.0 cm^3 / 1000 cm^3) = 0.005 mol. Since HCl and Ba(OH)2 neutralize each other in a 1:1 ratio, we need the same number of moles of HCl. Therefore, the volume of 0.200 mol/dm^3 HCl needed is (0.005 mol) / (0.200 mol/dm^3) = 0.025 dm^3, which is equivalent to 25.0 cm^3.
Is 12 mol HCl the same with 12 M HCl?
The Mole is a unit for amount of substance. Molarity, on the other hand, is a unit for concentration of a solution. 1M = 1 mol/dm^3.
How do you convert mol dm-3 to molL-1?
It is supposed to be mol/dm-3 Actually, 1 dm cube is the same as 1 litre. Therefore, there is no need of conversion. Both are the same.
What volume of a 0.10 mol dm-3 solution of HCl is required to react with 10.0 g of CaCO3?
2 HCl + CaCO3 = CaCl2 + CO2 + H2O Therefore it takes twice as many moles of HCl to react with CaCO3 Mol CaCO3 = mass / formulae weight = 10 / 40.08 + 14.01 + 3(16.00) = 10 / 102.09 = 0.0980 mol Mol HCl = 2 x 0.0980 = 0.1959 mol Volume HCl = mol / c = 0.1959 / 0.10 = 1.9591 L Volume required to react of HCl = 1.9591 L
23.74cm3 0.0147 mol dm-3NaOHare required to neutralize 25.00 cm3 of HCl calculate concentration of HCl acid in moldm-3and g dm-3?
HCl + NaOH ― NaCl + H2O nNaOH = c*v = 0.01470 * 23.74*10-3L = 3.49*10-4 mol 1 mol HCl reacts with 1 mol NaOH therefore nHCl = 3.49*10-4mol CHCL= nNaOH /VNaOH = 3.49*10-4 / 25.00*10-3 = 13.96 * 10-3 mol.dm-3
How do you calculate the pH of a solution of 0.001 mol dm HC1?
To calculate the pH of a 0.001 mol/dm³ HCl solution, you use the formula pH = -log[H+]. For HCl, since it is a strong acid, it dissociates completely into H⁺ ions. Thus, the concentration of H⁺ ions in a 0.001 mol/dm³ HCl solution is also 0.001 mol/dm³. Taking the negative logarithm of 0.001 gives a pH of 3.
What volume of the 0.200mole dm-3hcl is required to neutralize 25.0cm3 of 0.200 moldm-3 BaOH2?
To neutralize the Ba(OH)2, we need an equal number of moles of HCl. The molarity of Ba(OH)2 is 0.200 mol/dm^3, so the number of moles present in 25.0 cm^3 is (0.200 mol/dm^3) x (25.0 cm^3 / 1000 cm^3) = 0.005 mol. Since HCl and Ba(OH)2 neutralize each other in a 1:1 ratio, we need the same number of moles of HCl. Therefore, the volume of 0.200 mol/dm^3 HCl needed is (0.005 mol) / (0.200 mol/dm^3) = 0.025 dm^3, which is equivalent to 25.0 cm^3.
Is 12 mol HCl the same with 12 M HCl?
The Mole is a unit for amount of substance. Molarity, on the other hand, is a unit for concentration of a solution. 1M = 1 mol/dm^3.
How do you convert mol dm-3 to molL-1?
It is supposed to be mol/dm-3 Actually, 1 dm cube is the same as 1 litre. Therefore, there is no need of conversion. Both are the same.
What volume of a 0.10 mol dm-3 solution of HCl is required to react with 10.0 g of CaCO3?
2 HCl + CaCO3 = CaCl2 + CO2 + H2O Therefore it takes twice as many moles of HCl to react with CaCO3 Mol CaCO3 = mass / formulae weight = 10 / 40.08 + 14.01 + 3(16.00) = 10 / 102.09 = 0.0980 mol Mol HCl = 2 x 0.0980 = 0.1959 mol Volume HCl = mol / c = 0.1959 / 0.10 = 1.9591 L Volume required to react of HCl = 1.9591 L
How do you convert dm to mol?
It is supposed to be mol/dm-3 Actually, 1 dm cube is the same as 1 litre. Therefore, there is no need of conversion. Both are the same.
Calculate the H plus and OH- in the solution that contains 0.1 g HCl dissolved in 2.5 lit of solution?
Concentration of H+ can be written as [H+]. The square brackets denote "concentration of".Mr HCl = 36.5 g mol-1Moles HCl = 0.1 g / 36.5 g mol-1= 2.74 x 10-3 mol[HCl] = 2.74 x 10-3 mol / 2.5 L= 1.1 x 10-3 mol L-1 ( or mol dm-3)[H+] = 1.1 x 10-3 mol L-1[OH-] = (10-14) / (1.1 x 10-3) = 9.13 x 10-12 mol L-1pH = 2.96= 3.0
What is the pH of HCl?
well depending on the concentration of the solution but heres the equation to work it out: pH= -log[H+] so if the solution was a 1 mol dm^-3 HCl then: pH= -log(1) = 0 Hence 1 mol dm^-3 of HCl is VERY acidic Remember pH:0 is very acidic/ pH:7 in neutral/ and pH:14 is very basic
What is the molarity of a HCI solution that contains 10.0 g of HCI in 250 mL of solution?
To find the molarity, first calculate the number of moles of HCl using its molecular weight (36.46 g/mol). The number of moles is 10.0 g / 36.46 g/mol = 0.274 moles. Then, divide the number of moles by the volume of the solution in liters (250 mL = 0.25 L) to get the molarity: 0.274 moles / 0.25 L = 1.096 M. Thus, the molarity of the HCl solution is 1.096 M.
What is the hydrogen ion concentration in mol dm 3 when 1.50 g of sulfuric acid is dissolved in water to give 0.500 dm3 of solution?
6.17 x 10-3 mol dm-3
Substance with a pH of 3?
A solution with pH of 3 is one with a Hydrogen ion (H+) concentration of 0.001 mol dm-3. An example would be a 0.001 mol dm-3 Hydrochloric Acid (HCl) solution, because HCl is a strong acid which fully dissociates to H+ and Cl- ions in solution. However, most acids could be made into a solution of pH 3 at appropriate concentrations.
