Calculate the ph of the solution made by adding 0.50 mol of hobr and 0.30 mol of kobr to 1.00 l of water. the value of ka for hobr is 2.0ă—10â’9. express your answer numerically using two decimal places.

Here we have to get the the of the solution containing HOBr and KOBr.

The of the solution is 8.47

The mixture of HOBr and KOBr is a buffer solution. The of the buffer solution is determined by the Henderson equation which is:

= + log

Here [salt] i.e. KOBr is 0.30 mol/L and [Acid] is 0.50 mol/L.

The value of HOBr is 2×10⁻⁹.

We know, = -log

Or, = -log (2×10⁻⁹)

Or, = 8.698

On plugging the values:

= 8.698 + log

Or, = 8.698 - 0.221

Or, = 8.47

Thus the of the solution is 8.47.

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.