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Posted by on Wednesday, January 23, 2013 at 12:08pm.

calculate the pH of the solution when 25.0ml of 0.0920M HCL is titrated with 0.15ml, 23.0ml, 30.0ml of 0.10M NaOH

  • chemistry - , Wednesday, January 23, 2013 at 1:24pm

    Hopefully someone else comes along and checks this, but I'll give it a try. First convert all units of volume from mL to L and multiply them by their respective concentrations, which are given to you in molarity to find the number of moles. e.g. 25ml=0.025L*0.0920M gives you the number of moles of HCl. Subtract the number of moles of base from the number of moles of acid. If you have something left over, plug that number into the formula -log(H+)=pH. If you get 0, the reaction is a neutralization reaction and the pH is 7. if you get a negative value, that is the amount of base left over. Take the absolute value of that value and plug it into the formula -log(OH-)=pOH. This will give you the pOH. Subtract the pOH from 14 to give you the pH. Again I am not sure, but I hope this is correct. If this is, you should get a pH of 2.64, 7.00, and 10.8.

  • chemistry - , Wednesday, January 23, 2013 at 3:13pm

    mols HCl = 0.0920 x 0.025 = 0.0023
    mols NaOH = 0.100 x 0.00015 = 0.15E-5
    mols HCl remaining = 0.0023-1E-5 = 0.002285
    MHCl remaining = 0.002285 mols/(0.025+0.00015)L = 0.09085
    pH = -log(0.09085) = about 1.04 for 0.15 mL NaOH.

    Devron is correct for 23.0 mL 0.1M NaOH; the pH = 7.0

    For 30.0 mL of 0.1M NaOH, it's done the same way.
    30.0 x 0.1M NaOH = 0.003 mols NaOH
    25.0 x 0.0920 M HCl = 0.0023 mol HCl.
    Difference is 0.0007 mols NaOH.
    M NaOH = 0.007 mol NaOH/(0.025+0.030)L = 0.0007/0.055= 0.0127
    pOH = -log(OH^-)
    pOH = -log(0.0127)
    pOH = 1.89
    pH = 14-1.89 = about 12.1
    Check my work carefully, especially for typos. I don't like to work these things in mols, I prefer millimoles because there are fewer zeros to count.

  • chemistry - , Wednesday, January 23, 2013 at 3:30pm

    Sorry, I forgot to convert everything back to molarity before plugging back in the values.

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