What would the solution pH be if all the sulphite were destroyed by the second reaction, and then all the acetic acid that remained was destroyed by the first reaction?Discuss

1. The acid sulphite pulping process uses sulphurous acid, H2SO3, to attack the lignin that holds the wood fibres together. The fibres are then released and processed into paper products, and a hot solution containing acetic acid, sulphurous acid, and some larger organic molecules is generated as a waste.

This waste solution is partially evaporated and then condensed, with the smaller molecules being transferred into the condensate and the larger ones remaining behind in an organic-rich solution that can be burned to recover its energy content. The condensate can be treated in an anaerobic biological treatment process, but only if the pH is maintained near neutrality (pH 7).

When the process is operating successfully, the micro-organisms mediate the following reactions: CH3COOH + H2O CH4 + H2CO3 4H2SO3 + 3CH3COOH 4H2S + 6H2CO3

a. Find the pH of a solution of 5000 mg/L HAc and 300 mg/L H2SO3 as S (note units).

To do this generate a log C-pH diagram using Excel (with acetate and sulphite systems). Label all the lines and show on the diagram where the equilibrium pH is. You can use a charge balance to solve for pH.

b. What would the solution pH be if all the sulphite were destroyed by the second reaction, and then all the acetic acid that remained was destroyed by the first reaction?

Again, do this by generating a log C-pH diagram with Excel (with sulphide and carbonate systems). i.e. after rn 2: TOTSO3 = 0, TOTS = TOTSO3 initially TOTAc = initial –reacted TOTCO3 = that created from H2SO3 After rn

1: TOTAc =0 TOTS doesn’t change TOTCO3 = that from the 1st rn + that from 2nd rn Then set up problem and solve using log C-pH diagram

c. How much NaOH, in moles per litre, would be needed to bring the initial solution pHinto the range of good reactability, say, pH 7.0?

How much Na2CO3 would be required to accomplish the same result?

Comment on the relative requirements of these two bases, considering their relative strengths and the number of protons each can accept. i.e. use CB from a and solve for Na+ With Na2CO3 must include the CO3 system in CB, noting that [Na+] = 2TOTCO3 –sub into CB and calculate TOTCO3

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