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How do you prepare 0.5L of 65 percent Nitric acid from 69.3 percent of Nitric acid?
Wiki User
∙ 12y ago
It depends on if the Assay is in wt / wt or wt / L because if that's the case you will need to find the density of the 67-70% HNO3 at room temp.
But basically you used the equation:
C1V1 C2V2
where
C1 initial concentration
V1 the amount of solution you will need
C2 final concentration
V2 final volume
I would take the average of the assay for C1, so
(67+70) / 2 68.5
you need to know how much of the solution you want for example 3 L
then you plug it in, this is also assuming that the initial concentration is in wt / L
C1 * V1C2 * V2
68.5% * V12% * 3L
rearrange to form:
V1 (2% * 3L) / 68.5% 0.0876L
which you can change into Wt / Wt with the density, or if you need mL just convert with ( 1000mL / 1L )
Wiki User
∙ 11y ago
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∙ 12y agoCarefully.
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How do you calculate the number of moles of NaI in 50.0mL of a 0.400M solution?
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A buffer is prepared by mixing 50.0 ml of 0.050 m sodium bicarbonate and 10.7 ml of 0.10 m naoh what is the pH for carbonic acid ka1 equals 4.5 x 10 -7 and ka2 equals 4.7 x 10 -11?
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.05l equals how many ml?
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50 ml of point05M HCl to 150 ml of point10 M HNO3 what are the concentrations of all species in this solution?
I'll assume when you say species, you are referring to the ions of each compound, so we are looking for the concentration of H+, Cl-, and NO3-.To begin, we have 0.2L of total solution (.05L HCl + .15L HNO3).I'll do Cl- first: You have .05L of 0.05mol/L HCl. H and Cl are in 1:1 ratio, so you can say you have 0.05L of 0.05mol/L Cl-.0.05L * 0.05mol/L=0.0025mol Cl- divide this by the total volume, 0.0025 mol/.200L=0.0125M Cl-NO3-: .150L*0.1M NO3-= 0.015mol NO3-/0.200L=0.075M NO3-H+ occurs in both acids so you have to add both amounts to find the concentrations. We know that for HCl, H has a 1:1 ratio with Cl, so if there is 0.0025mol Cl- there must be 0.0025mol of H+.Same with HNO3, there must be 0.015mol H+.Add these together: 0.0025mol+0.015mol= 0.0175mol H+ in total.Divide by the total volume to get concentration: 0.0175mol/.200L= 0.0875M H+
