Question

Using a 0.30 M phosphate buffer with a pH of 6.5, you add 0.80 mL of 0.55 M NaOH to 41 mL of the buffer. What is the new pH of the solution?

Answer 1

As the pH is around the pka₂ value (7.21) , you have the species H₂PO₄^- and HPO₄^2- (mainly).

First you need to obtain the initial concentrations, these can be obtained with the Henderson-Hasselbalch equation:

(based pH = pka + log Cacid

that with the given values can be written as:

THPOR? ] 6.5=7.21+ 100 y THz Pow]

the other equation that we need is the total concentration:

THPOR] +[Hz PO, ]=0.30M

isoltaing [HPO₄^2- ] we have:

CHPO,?] =0,30M-CH2 pou ]

this value can be substituted in the Henderson-Hasselbalch equation to obtain [H₂PO₄^-]:

6.S= 1.2| + . CHeo,-1 Tog TH₂PO4 6.5=7.21+ log 0.30M-[Mq Pox-1 [H₂PO4] -0.7 = loa 5.35 -CH2PO, 1 1. The Pou - - 10-0.71 ] 0.30M-[H₂ PO [H2Pou - 0.30M --1 [H₂PO4 0.1950 -

0. 30M 1.1950- [H₂ PO. (H₂PO4 ] = 0.2510 M.

Now we can know [HPO₄^2- ]:

CHPO.?]=0.30M-[Hz pou] =0.30M-0.2510M [HPO ? - ] = 0.0490M.

Next, we have to know the number of moles in the given volume of buffer:

mupoz + (41m2) ) = 0.0020 mol Maior = lume ) ( aste boek ) - 0.0103 wol MzPO 0.2510mol 00

We can calculate the final concentrations if we know the number of moles of base that will react, this is obtained as follows:

Mhuset (0.80ml) Kalo mal ) - 44410*mo! VRE WO 1,000mL

So, knowing that each mol of the buffer acid will react with one mole of added base to form one mol of buffer base:

nu pozFinal = 0.0020molt 4.4x10^me l = 0.0024 met Nhz pour final = 0.0103 mol - 4.4x10"mol = 0.0099 mol

In the Henderson-Hasselbalch equation we can write number of moles instead of concentrations because they are in the same volume:

Topou?-] PO pH = 7.21+ log Thement pH=7.21 +log (-0.0024 pH = 6.59 0,0099