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.
The pH of a 2M HNO3 solution is approximately 0. This is because nitric acid (HNO3) is a strong acid that fully dissociates in water to release H+ ions, resulting in a highly acidic solution with a low pH value.
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 )
Since HNO3 is a strong acid, it completely dissociates in solution. HNO3 -> H+ + NO3-. Therefore, the concentration of H+ ions is the same as the concentration of the HNO3 solution, 0.0067M. pH = -log[H+] = -log(0.0067) ≈ 2.18. pOH = 14 - pH ≈ 11.82.
The pH of a 1.45M HNO3 solution is approximately 0.14. This is because nitric acid is a strong acid that dissociates completely in water to yield H+ ions.
The pH of a 0.6 M HNO3 solution is approximately 0.23. This is because nitric acid is a strong acid that completely ionizes in solution, resulting in a high concentration of H+ ions that lower the pH.
pH = pKa - log([HA]/[A])[HA] = concentration weak acid NH4+ and [A] = concentration of weak base NH3 Before HNO3 addition [HA]= 6.0 mmol NH4+ in 100 ml => after: [HA]= 6.0+5.3 mmol NH4+ in 101 mlBefore HNO3 addition [A]= 11.5 mmol NH3 in 100 ml => after: [HA]= 11.5-5.3 mmol NH3 in 101 mlThus: (notice that pH of buffer is independent from a large range of dilution)pHbefore = 9.24- log10(6.0/100 / 11.5/100) = 9.24 - log10(6/11.5) = 9.52and:pHafter = 9.24 - log10(11.3/101 / 6.2/101) = 9.24 - log10(11.3/6.2) = 8.98
The pH of a 2M HNO3 solution is approximately 0. This is because nitric acid (HNO3) is a strong acid that fully dissociates in water to release H+ ions, resulting in a highly acidic solution with a low pH value.
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 )
Since HNO3 is a strong acid, it completely dissociates in solution. HNO3 -> H+ + NO3-. Therefore, the concentration of H+ ions is the same as the concentration of the HNO3 solution, 0.0067M. pH = -log[H+] = -log(0.0067) ≈ 2.18. pOH = 14 - pH ≈ 11.82.
The pH of a 1.45M HNO3 solution is approximately 0.14. This is because nitric acid is a strong acid that dissociates completely in water to yield H+ ions.
The pH of a 0.6 M HNO3 solution is approximately 0.23. This is because nitric acid is a strong acid that completely ionizes in solution, resulting in a high concentration of H+ ions that lower the pH.
- log(0.01 M HNO3) = 2 pH =====
pH = -log[H+] pH = -log[1.6 × 10-3] pH = 2.8
the lower the pH the stronger the acid
The pH value of those solutions depends on the concentration and the temperature. Generally, solutions with higher concentrations of acids have lower pH values.
The concentration of HNO3 in a solution with pH 3.4 is approximately 3.98 x 10^-4 M. This is calculated using the formula pH = -log[H+], where [H+] is the hydrogen ion concentration in mol/L. For nitric acid (HNO3), one mole of HNO3 produces one mole of H+ in solution.
The pH of HNO3 is about 0.1. It is a strong acid, and when dissolved in water, it completely dissociates into H+ and NO3- ions. This high concentration of H+ ions results in a low pH value close to zero.