What is an equation of the line that is perpendicular to y and ndash 4 2(x and ndash 6) and passes through the point ( and ndash3 and ndash5)?

In an exactly.200M solution of benzoic acid a monoprotic acid the H plus 3.55 and times 10 and ndash3M. What is the Ka for benzoic acid 6.3 and times 10 and ndash5 1.77 and times 10 and ndash2 8.88 an?

The correct Ka value for benzoic acid is 6.3 x 10^-5. This can be calculated using the equation Ka = [H+][C6H5COO-] / [C6H5COOH]. Plugging in the values given ([H+] = 3.55 x 10^-3 M, [C6H5COOH] = 0.200 M) gives Ka = (3.55 x 10^-3)(3.55 x 10^-3) / (0.200) = 6.3 x 10^-5.