Calculate the pH at the point in the titration of 25.00 mL 0.108 M HNO2 at which 10.00 mL 0.162 M NaOH have been added. For HNO2, Ka=5.1 x 10^-4 and:
HNO2 + OH- -----> H2O + NO2-
I know that HNO2 is a weak acid and that 5.1 x 10^-4=[NO2-]/[HNO2][OH-]. NaOH is a strong acid . Not sure how to go from here. Thanks.
oops..I meant NaOH is a strong base....
HNO2 + NaOH ==> NaNO2 + H2O
moles HNO2 = M x L = 0.025 x 0.108 = 0.00270 moles HNO2 initially.
moles NaOH added = 10mL x 0.162 = 0.00162
moles HNO2 remaining after reacction = 0.00270-0.00162 =0.00108.
M HNO2 = moles/L (L = 25 mL + 10 mL)
So this problem becomes determine the pH of a 0.00108/0.035)M soln of HNO2. Set up an ICE chart, solve for H^+ and convert to pH.
what does ICE stand for? thanks.
To solve this problem, we need to calculate the concentration of HNO2 and OH- at the point when 10.00 mL of 0.162 M NaOH has been added to 25.00 mL of 0.108 M HNO2.
First, let's calculate the number of moles of NaOH that have been added:
Moles of NaOH = Volume (in liters) x concentration
= 0.010 L x 0.162 M
= 0.00162 moles
Next, we need to determine the remaining moles of NaOH after it reacts with HNO2. The reaction between HNO2 and NaOH is a 1:1 ratio, meaning one mole of HNO2 reacts with one mole of NaOH. Therefore, the number of moles of HNO2 that reacted with NaOH is also 0.00162 moles.
Now, we can subtract the moles of HNO2 that reacted with NaOH from the initial moles of HNO2 to find the moles of HNO2 remaining:
Moles of HNO2 remaining = Initial moles of HNO2 - Moles of HNO2 reacted with NaOH
= (Volume x concentration) - 0.00162 moles
= (0.025 L x 0.108 M) - 0.00162 moles
= 0.0027 moles
To calculate the concentration of HNO2 remaining, divide the moles of HNO2 remaining by the total volume of the solution:
Concentration of HNO2 remaining = Moles of HNO2 remaining / Total volume
= 0.0027 moles / (0.010 L + 0.025 L)
= 0.0027 moles / 0.035 L
= 0.077 M
Now, we need to calculate the concentration of NO2- produced by the reaction between HNO2 and NaOH. Since the reaction is a 1:1 ratio, the concentration of NO2- is equal to the concentration of HNO2 remaining:
Concentration of NO2- = Concentration of HNO2 remaining
= 0.077 M
Finally, let's calculate the OH- concentration using the given equilibrium equation:
Ka = [NO2-] / [HNO2][OH-]
Substituting the values into the equation:
5.1 x 10^-4 = 0.077 M / (0.077 M)([OH-])
Simplifying the equation, we get:
[OH-] = 5.1 x 10^-4 / 0.077 M
= 0.006623 M
To calculate the pH, we need to calculate the pOH first:
pOH = -log10([OH-])
= -log10(0.006623)
Finally, we can calculate the pH using the equation:
pH = 14 - pOH
= 14 - (-log10(0.006623))
Now, you can use a calculator to find the pH.