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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 )
If 0.630 grams of HNO3 molecular weight 63.0 are placed in a liter of distilled water at 25 degree Celsius what will be the pH of the solution?
2
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 )
If 0.630 grams of HNO3 molecular weight 63.0 are placed in a liter of distilled water at 25 degree Celsius what will be the pH of the solution?
2
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.
If hno3 is added to water how does pH- change?
If nitric acid (HNO3) is added to water, it decreases the concentration of hydroxide ions in solution. This is because nitric acid semi-strongly dissociates in water, following this chemical reaction: HNO3(aq) + H2O(l)-->NO3-(aq) + H3O+(aq) The hydronium ions that are created in this reaction then react quickly with the hydroxide ions in the water, as shown in this chemical equation: H3O+(aq) +OH-(aq) --> 2H2O(l) This results in fewer hydroxide ions existing in solution.
What is the pH of 2M HNO3?
pH= -log [H+] = -log [1] = 0
Which measurement is the largest to use 145m or 145km?
145m or 145km
How many moles of hno3 are needed to prepare 5.0 of a 2.0 m solution of hno3?
10
Suppose you wanted to produce an aqueous solution of pH 8.75 by dissolving potassium nitrate in water what is the molarity?
Not likely. Potassium nitrate, KNO3 is the salt of a strong acid (HNO3) and a strong base (KOH). Thus, the pH of a solution of KNO3 will be very close to pH=7, depending really on the pH of the water used to make the solution. There is no way to get it to pH = 8.75. You'd need to used something like potassium acetate, the salt of a strong base and a weak acid.
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.
