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1.22Molarity (multiply) 0.076Liters = 0.09272Moles needed to neutralize. 0.09272Moles (divided by) 0.125Liters = 0.74176 Molarity HCl has a Molarity of 0.74176
Molarity = moles of solute/liters of solution or, for our purposes moles of solute = liters of solution * Molarity moles of AgNO3 = 0,50 liters * 4.0 M = 2.0 moles of AgNO3 needed --------------------------------------
0.564 M
Firstly, write out and balance your equation (as always!); this'll also require calculating the Molarity (mol of solute/ ml of solution). Next, you should already have an indicator (litmus strips or even red cabbage juice will work). Then add exactly the amount needed to neutralize the intended reagents. If you're doing the work on paper only, ignore this part: However, have you considered baking soda as opposed to NaOH? It's less dangerous.
Balanced equation first, last and always! Na2CO3 + 2HCl - > 2NaCl + CO2 + H2O 2.5 g Na2CO3 ( 1 mole Na2CO3/105.99 g)(2 mole HCl/1 mole Na2CO3) 0.04717 moles HCl -------------------------Now, Molarity = moles of solute/liters of solution or, for our purposes liters of solution = moles of solute/Molarity Liters HCl = 0.04717 moles HCl/0.60 M HCl = 0.0786 liters (1000 milliliters/1 liter) = 78. 6 milliliters HCl solution needed
1.22Molarity (multiply) 0.076Liters = 0.09272Moles needed to neutralize. 0.09272Moles (divided by) 0.125Liters = 0.74176 Molarity HCl has a Molarity of 0.74176
How many grams of KHP are needed to exactly neutralize 36.7 mL of a 0.328 M barium hydroxidesolution
Molarity = moles of solute/liters of solution or, for our purposes moles of solute = liters of solution * Molarity moles of AgNO3 = 0,50 liters * 4.0 M = 2.0 moles of AgNO3 needed --------------------------------------
What is the molarity of an HCl solution if 43.6 mL of a 0.125 M KOH solution are needed to titrate a 25.0 mL sample of the acid according to the equation below?
0.564 M
22
0.0532
Molarity = moles of solute/volume of solution ( so, not a great molarity expected ) 4.60 grams H2SO4 (1mol H2SO4/98.086g) = 0.0469 moles/450ml = 1.04 X 10^-4 Molarity.
Firstly, write out and balance your equation (as always!); this'll also require calculating the Molarity (mol of solute/ ml of solution). Next, you should already have an indicator (litmus strips or even red cabbage juice will work). Then add exactly the amount needed to neutralize the intended reagents. If you're doing the work on paper only, ignore this part: However, have you considered baking soda as opposed to NaOH? It's less dangerous.
0.0932 L
2
Balanced equation first, last and always! Na2CO3 + 2HCl - > 2NaCl + CO2 + H2O 2.5 g Na2CO3 ( 1 mole Na2CO3/105.99 g)(2 mole HCl/1 mole Na2CO3) 0.04717 moles HCl -------------------------Now, Molarity = moles of solute/liters of solution or, for our purposes liters of solution = moles of solute/Molarity Liters HCl = 0.04717 moles HCl/0.60 M HCl = 0.0786 liters (1000 milliliters/1 liter) = 78. 6 milliliters HCl solution needed