A weak base BOH was titrated against a strong acid. The pH at one-fourthequivalence point was 9.24. Enough strong base was now added (6 m.e.) to completely convert the salt. The total volume was 50 mL. Find the pH at this point. 6 m.e. of the strong base, added, is used to convert the salt to the weak base. Thus before the addition of the strong base, m.e. of the salt and the base were 6 and 18 respectively. As 6 m.e. of the strong base shall combine with the same number of m.e. of the salt to produce 6 m.e. of BOH, total m.e. of BOH = 6 + 18 = 24 and thus molarity = 24/50 M. Now using K_b value, calculate the pH.

A weak base BOH was titrated against a strong acid. The pH at one-four

1.	A weak base BOH was titrated against a strong acid. The pH at one-fourthequivalence point was 9.24. Enough strong base was now added (6 m.e.) to completely convert the salt. The total volume was 50 mL. Find the pH at this point. 6 m.e. of the strong base, added, is used to convert the salt to the weak base. Thus before the addition of the strong base, m.e. of the salt and the base were 6 and 18 respectively. As 6 m.e. of the strong base shall combine with the same number of m.e. of the salt to produce 6 m.e. of BOH, total m.e. of BOH = 6 + 18 = 24 and thus molarity = 24/50 M. Now using K_b value, calculate the pH.
Answer» SOLUTION :`(14-9.24) = pK_b + LOG((1//4)/(3//4)) `, CAL `K_b` 11.2