1. Sulfuric acid is diprotic that will completely ionize to give two protons whe
ID: 1025246 • Letter: 1
Question
1. Sulfuric acid is diprotic that will completely ionize to give two protons when placed in with a strong base like potassium hydroxide.
Write a balanced equation for the reaction.
How many mL of 0.24 M potassium hydroxide are needed to neutralize 18.3 mL of 0.56 M sulfuric acid?
2. Calculate the pH of a solution that results from 25.0 mL of 0.525 M NaOH being mixed in the same beaker as 75.0 mL of a 0.355 M HCl solution.
3. Picture that 200. mL of a buffered solution is made by adding 100. mL of 0.10 M HNO2 to 100. mL of 0.10 M Ca(NO2)2.
a)Write the equation for the buffer system, leaving out any spectator ions.
b)If 5.0 mL of 0.1 M NaOH are added, what does the base react with in the buffer system? (Write it as an equation: NaOH + what from the ‘buffer system yields’...).
c)Predict whether the buffer system that’s been made will be a high pH or a low pH buffer system? Explain. (HINT: remember acid/base properties of salt solutions)
Explanation / Answer
H2SO4(aq) + 2KOH(aq) -------------> K2So4(aq) + 2H2O(l)
no of moles of H2SO4 = molarity * volume in L
= 0.56*0.0186
= 0.010416moles
1 mole of H2So4 react with 2 moles of KOH
0.010416 moles of H2So4 react with = 2*0.010416/1 = 0.020832moles KOH
no of moles of KOH = molarity * volume in L
0.02083 =0.24* volume in L
volume in L = 0.02083/0.24 = 0.0867L = 86.7ml
2.
HCl NaOH
MA = 0.355M MB = 0.525M
VA = 75ml VB = 25ml
M = MAVA - MBVB/VA+ VB
= 0.355*75-0.525*25/75+25
= 13.5/100 = 0.135M
[H^+] = M = 0.135M
PH = -log[H^+]
= -log0.135
= 0.8696
3. no of moles of HNO2 = molarity * volume in L
= 0.1*0.1 = 0.01moles
no of moles of Ca(NO2)2 = molarity * volume in L
= 0.1*0.1 = 0.01moles
PH = Pka + log[Ca(NO2)2]/[HNO2]
= 3.15 + log0.01/0.01
= 3.15
no of moles of NaOH = 0.1*0.005 = 0.0005 moles
no of moles of HNO2 after addition of NaOH = 0.01-0.0005 = 0.0095 moles
no of moles of Ca(OH)2 after addition of NaOH = 0.01+0.0005 = 0.0105 moles
PH = PKa + log[Ca(NO2)2]/[HNO2]
= 3.15 + log0.0105/0.0095
= 3.15+0.04346 = 3.1935
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