Where did the 0.40 and 0.60 come from? Please show work!! QS: Use activity coeff

Question

Where did the 0.40 and 0.60 come from? Please show work!!

QS: Use activity coefficients to calculate the pH after 10.0 mL of 0.100 M trimethylammonium bromide was titrated with 4.0 mL 0.100 M NaOH.

answer:

(CH3)3NH++ OH- (CH3)3N + H2O Initial m 00 Final mmol:0.60 0.40 0.40 First find the ionic strength: l(CH3)3NH+] = 0.60 mmol/ 14.0 mL = 0.04286 M [Br] = 1.00 mmol/ 14.0 mL = 0.071 43 M [Na+] 0.40 mmol/ 14.0 mL = 0.028 57 M pH = pKa + log [BH+/BH+ pH = 9.799 + log (00429)(0.80)- (0.028 61.009.72 In the previous calculation, we used the size of (CH3)3NH+ (400 pm) and the activity coefficient interpolated from Table 8-1.

Explanation / Answer

4.0 mL 0.100 M NaOH

By definition of molarity, we know, 1000 ml 1M NaOH = 1 mol NaOH

so, 1 ml, 1M = 1 mmol of NaOH

No, 4 ml 1M = 4 mmol of NaOH

And for 4 ml 0.100 M = 0.40 mmol of NaOH is the initial concentration of NaOH

It is given the initial concentration of (CH3)3NH+ ion is 1.0 mmol

So, after reaction with 0.4 mmol of NaOH remaining (CH3)3NH+ ion concentration = (1-0.4) mmol = 0.60 mmol

0.40 mmol of NaOH and 0.060 mmol of (CH3)3NH+ ions, these are two values used in the calculation.

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