potassium chloride
If the solution is not a buffer, the HCl will react with the solution to form a product.
(7 mL)(X M HCl) = (27.6 mL)(0.170 M NaOH)7X = 4.692X = 0.7 M HCl==========
0.2 N HCl solution means 0.2 equivalents of HCl dissolved in 1 litre of water. Normality = Molarity x n-factor => Molarity =Normality/n-factor=0.2/1=0.2 M 0.2 moles should be present in 1 litre of solution. 0.2moles =0.2 x 36.5 = 7.3 grams of HCl =>Dissolve 7.3 grams of HCl in 1 litre if water to get a 0.2N solution.
40 ml of NaOH contains 0.04 L * 3.5 M = 0.14 mole of NaOH Since NaOH donates 1 OH you will also have 0.14 mole of OH- in solution. This can be neutralised with an equal amount of H+. HCl can donate 1 H+, so you need an equal amount of H+ to neutralise the OH-. So you need 0.14 mole of the HCl. 55 ml has 0.14 mole HCl. So the molarity is: 0.14 mole / 0.055 L = 2.54 M
To neutralise the acid (HCl).
potassium chloride
Quantitative observations involve numbers. An example would be the volume of 0.1M HCl needed to neutralise 25 ml of sodium hydroxide solution.
If the solution is not a buffer, the HCl will react with the solution to form a product.
(7 mL)(X M HCl) = (27.6 mL)(0.170 M NaOH)7X = 4.692X = 0.7 M HCl==========
If it is only HCL, yes.
yes it is
0.2 N HCl solution means 0.2 equivalents of HCl dissolved in 1 litre of water. Normality = Molarity x n-factor => Molarity =Normality/n-factor=0.2/1=0.2 M 0.2 moles should be present in 1 litre of solution. 0.2moles =0.2 x 36.5 = 7.3 grams of HCl =>Dissolve 7.3 grams of HCl in 1 litre if water to get a 0.2N solution.
40 ml of NaOH contains 0.04 L * 3.5 M = 0.14 mole of NaOH Since NaOH donates 1 OH you will also have 0.14 mole of OH- in solution. This can be neutralised with an equal amount of H+. HCl can donate 1 H+, so you need an equal amount of H+ to neutralise the OH-. So you need 0.14 mole of the HCl. 55 ml has 0.14 mole HCl. So the molarity is: 0.14 mole / 0.055 L = 2.54 M
HCl is ionozed in aq solution HCl + H2O = H3O(+ CHARGE) + Cl (- CHARGE) HCl FORMULA WILL REAMAIN HCl OT WILL BOT CHANGE
HCl is a strong acid. Therefore, it can be expected to fully dissociate in aqueous solution, yielding one hydrogen ion and one chloride ion per molecule. The concentration of the hydrogen ion should thus be the same as the initial concentration of the HCl. Therefore, a 0.10M HCl solution has an H+ concentration of 0.10M. By the equation pH=-log[H+], the pH of this solution is 1.
Molarity = moles of solute/liters of solution Molarity = 0.597 moles HCl/0.169 liters = 3.53 M HCl ------------------