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Normality= molarityX total POSITIVE oxidation number of solute Solvent=substance present in the greatest amount Solute= all other substances in the solution +1 (1 na+) is the total POSITIVE oxidation number of NaOH 2.0X1= 2.0 N=2.0
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∙ 16y ago
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∙ 13y agopH = -log [H+ ion concentration]
HNO3 is a strong acid, and because it is strong you can just plug 2.0M in for the H+ ion concentration and solve.
If it were a strong base, you would take the answer you got for the above problem as your pOH instead of pH. pOH+pH=14. This method only works for strong acids or bases.
Wiki User
∙ 12y agoWiki User
∙ 9y agoSince each NaOH molecule contains only one hydroxide ion, the normality of NaOH is equal to its molarity. Therefore, 1.0 N of NaOH is the same as 1 M.
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∙ 10y agoNaOH 1.0 N (normal) is standardised at 1.0 mole per Liter
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∙ 12y ago12
Anonymous
In the following reaction calculate and find the normality of the solution when 1.0 M H3PO4 reacts with NaOH?
H3AsO4 + 2NaOH → Na2HAsO4 + 2H2O
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What is the pH of a 0.01 M solution of the acid HNO3 in water?
- log(0.01 M HNO3) = 2 pH =====
What is the pH of a 1.6 10-3 M HNO3 solution?
pH = -log[H+] pH = -log[1.6 × 10-3] pH = 2.8
What is the pH of 15 mL of 0.0045 M HNO3?
pH = - log10 [H+], where [H+] is the molar concentration of hydrogen ions. HNO3 is a strong acid and dissociates completely in water so a 5 M solution of HNO3 would have a concentration of hydrogen ions of 5M also. So, pH = -log10[5] = -0.699 which indicates an extremely strong acid.
What is the pH of a solution that contains 1.32 grams of nitric acid dissolved in 750 milliters of water?
Two steps. Find molarity of nitric acid and need moles HNO3.Then find pH. 1.32 grams HNO3 (1 mole HNO3/63.018 grams) = 0.020946 moles nitric acid ------------------------------------- Molarity = moles of solute/Liters of solution ( 750 milliliters = 0.750 Liters ) Molarity = 0.020946 moles HNO3/0.750 Liters = 0.027928 M HNO3 ----------------------------------finally, - log(0.027928 M HNO3) = 1.55 pH ==========( could call it 1.6 pH )
What is the pH of the solution formed by completely neutralizing 50 milliliters of 0.1 m hno3 with 50 milliliters of 0.1 m naoh at 298 k?
its 7, seven is neutral.
What is the pH of a 0.01 M solution of the acid HNO3 in water?
- log(0.01 M HNO3) = 2 pH =====
What is the pH of a 1.6 10-3 M HNO3 solution?
pH = -log[H+] pH = -log[1.6 × 10-3] pH = 2.8
What is the pH of 15 mL of 0.0045 M HNO3?
pH = - log10 [H+], where [H+] is the molar concentration of hydrogen ions. HNO3 is a strong acid and dissociates completely in water so a 5 M solution of HNO3 would have a concentration of hydrogen ions of 5M also. So, pH = -log10[5] = -0.699 which indicates an extremely strong acid.
What is the pH of a solution that contains 1.32 grams of nitric acid dissolved in 750 milliters of water?
Two steps. Find molarity of nitric acid and need moles HNO3.Then find pH. 1.32 grams HNO3 (1 mole HNO3/63.018 grams) = 0.020946 moles nitric acid ------------------------------------- Molarity = moles of solute/Liters of solution ( 750 milliliters = 0.750 Liters ) Molarity = 0.020946 moles HNO3/0.750 Liters = 0.027928 M HNO3 ----------------------------------finally, - log(0.027928 M HNO3) = 1.55 pH ==========( could call it 1.6 pH )
What is the pH of the solution formed by completely neutralizing 50 milliliters of 0.1 m hno3 with 50 milliliters of 0.1 m naoh at 298 k?
its 7, seven is neutral.
How many moles of hno3 are needed to prepare 5.0 of a 2.0 m solution of hno3?
10
What is the concentration of HNO3 in a solution with pH3.4?
1/103.4= 4.0 x 10 -4 M HNO3==============
What would be pH of a 0.00884 M solution of HNO3?
HNO3 is a strong acid, which means it dissociates completely. This means you don't have to set up an equilibrium scenario; you can just go with the given molarity as also being the concentration of hydrogen ions [H+]. So, pH = -log(0.00884), which is about 2.05.
What is the molarity of an naoh solution if 4.37ml is titrated by 11.1 ml of 0.0904 m hno3?
0.289 Moles is the molarity of an NaOH solution if 4.37 ml is titrated by 11.1 ml of 0.0904m hno3.
The pH of a 0.01 M solution of HCl in water would be?
pH = - log[H+] so a 0.01 M solution of HCl has, pH= 2
What is the pH of a 0.001 M of HCl solution?
its PH is 3
What is the molarity of a solution dissolve 0.31 grams of HNO3 in 300ml of water?
Molarity = moles of solute/Liters of solution ( get moles of HNO3 and 300 ml = 0.300 Liters ) 0.31 grams Nitric acid (1 mole HNO3/63.018 grams) = 0.004919 moles HNO3 Molarity = 0.004919 moles HNO3/0.300 Liters = 0.0164 M HNO3
