Using a 0.25 M phosphate buffer with a pH of 6.2, you add 0.70 mL of 0.50 M HCl to 49 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)
Ph= PKa+log[ concentration of buffer/ concentration of acid]
6.2 =6.86+ log[phsophate/ acid]
6.2-6.86= log [0.25/acid]
-0.66 =log [0.25/acid]
0.25/acid=10^(-0.66)
acid =0.25/0.218= 1.146M
moles of buffer in 49 ml = 0.25*49/1000 = 0.01225
moles of acid added = 0.5*49/1000 =0.0245
volume of the mixture= 49+0.7 =49.7 ml
Concentration of buffer = 0.01225/(49.7/1000)= 0.246278M
Concentration of acid = 0.0245*1000/49.7= 0.4929 M
Now pH= pka+ log [ 0.24627/0.4929]
pH= 6.86-0.30138=6,55
Get Answers For Free
Most questions answered within 1 hours.