The definition of pH is the -log[H+], which basically means the negative common log (base 10) of the concentration of hydronium ions in solution. Since nitric acid is a strong monoprotic acid (one H+) it makes the problem so much easier since there is no acid equilibrium to deal with. Since there is one H+ for every acid molecule thus the [H+], or concentration of H+, is equal to the concentration of the acid, which is given to be 0.2004M. Thus, pH = -log[H+] = -log(0.2004) = (with sig figs) 0.6981 pH, something you really don't want to stick your hand in, but not as bad as some acid I've seen (yay, negative pH!).
- log(0.200 M HNO3)
= 0.700 pH
========
- log(0.01 M HNO3) = 2 pH =====
pH = -log[H+] pH = -log[1.6 × 10-3] pH = 2.8
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.
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 )
its 7, seven is neutral.
- log(0.01 M HNO3) = 2 pH =====
pH = -log[H+] pH = -log[1.6 × 10-3] pH = 2.8
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.
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 )
its 7, seven is neutral.
10
1/103.4= 4.0 x 10 -4 M 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.
0.289 Moles is the molarity of an NaOH solution if 4.37 ml is titrated by 11.1 ml of 0.0904m hno3.
pH = - log[H+] so a 0.01 M solution of HCl has, pH= 2
its PH is 3
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