Aqueous sulfurous acid (h2so3) was made by dissolving 0.200 l of sulfur dioxide gas at 19°c and 745 mmhg in water to yield 500.0 ml of solution. the acid solution required 16.9 ml of sodium hydroxide solution to reach the titration end point. what was the molarity of the sodium hydroxide solution?
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Ответ:
The molarity of the sodium hydroxide solution is 0.97 M
Explanation:
Step 1: Balanced reaction of H2SO3
SO2 + H2O → H2SO3
Step 2= Data given
Volume of SO2 = 0.2L
Temperature of SO2 = 19°C = 292.15 Kelvin
Pressure = 745 mmHg = 0.980263 atm
Via the ideal gas law we have this formula:
PV = nRT
with P = pressure of the gas = 0.980263 atm
with V = volume of the gas = 0.2L
with n = number of moles = TO BE DETERMINED
with R = Gas constant = 0.082057 L atm K−1 mol−1
Step 3: Calculating number of moles of SO2
0.980263 * 0.2 = n * 0.082057 * 292.15
n = 0.0082 moles
Step 4: Calculating number of moles of H2SO3
Since the ratio of the balanced equation is 1:1 for SO2 and H2SO3 this means that there is also 0.0082 moles of H2SO3
Step 5: Calculating moles of NaOH
The second reaction is H2SO3 + 2 NaOH → Na2SO3 + 2H2O
This means for 1 mole of H2SO3 consumed there is needed 2 moles of NaOH
So, for 0.0082 moles of H2SO3 there is needed 0.0164 moles of NaOH
Step 6: Calculating molarity of NaOH
The molarity of the NaOH solution can be calculated as:
Molarity = Moles / volume
Molarity = 0.0164 moles / 0.0169 L
Molarity = 0.97 M
The molarity of the sodium hydroxide solution is 0.97 M
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