Chemistry Problem--molarity/pH?

For most weak acid solutions, [H+] = sqrt (Ka x [HA]o) where [HA]o is the initial acid molarity.

[H+] = 10^-pKa = 10^-4.6 = 2.5 x 10^-5

[H+] = sqrt ((Ka)([HA]o))

2.5 x 10^-5 = sqrt ((Ka)(0.36)) . . .square both sides

6.3 x 10^-10 = 0.36 Ka

Ka = 1.8 x 10^-9

i choose to preface my answer by way of telling u that u can email me with diverse questions. it is not hassle-free to respond to all 3 questions, yet ill attempt. a million. we see that H3O+ is the best acid, via fact water is an exceedingly very susceptible base. thye conjugate of a sturdy acid is a susceptible base. the extra advantageous the acid, the weaker its conjugate. the extra advantageous the backside, the weaker its conjugate. 2. i comprehend for fantastically very just about specific that B(OH)3 is an ACID. in lots of situations, all nonmetal OH's are acids. if u rearrange this formulation, u get H3BO3. i will say D. 3. to discover the pH of a susceptible acid, u take the sq. root of Ka x Molarity. via fact all molarities are a million, take the sq. root of all of the Ka's and take the unfavorable log of what u have been given. and the closest one to 7 is the winner. it sticks out via fact the NH4+ i desire this facilitates.