Question

To calculate ph of a buffer of 0.105 m nahco3 and 0.125m na2co3

Answer 1

pKa (HCO3-) = 10.33

First case:
pH = pKa + log [CO3^2-]/[HCO3-] = 10.33 + log (0.350/0.220) = 10.33 + 0.20 = 10.53

Second case:
[CO3^2-] = 0.19 M * 75 mL / 140 mL = 0.102 M
[HCO3-] = 0.17 M * 65 mL / 140 mL = 0.079 M

pH = 10.33 + log (0.102/0.079) = 10.33 + 0.11 = 10.44

Answer 2

Hey dear,

● Answer -
pH = 6.42

● Explanation -
# Given -
[CO3--] = m(Na2CO3) = 0.125
[HCO3-] = m(NaHCO3) = 0.105

# Solution -
Buffer equation for Na2CO3 & NaHCO3 is given as -
HCO3- → H+ + CO3--

Acid constant is given by -
Ka = [H+][CO3--] / [HCO3-]
[H+] = Ka[HCO3-] / [CO3--]
[H+] = 4.5×10^-7 × 0.105 / 0.125
[H+] = 3.78×10^-7

pH of solution is calculated by -
pH = -log[H+]
pH = -log(3.78×10^-7)
pH = 6.42

Therefore, pH of buffer solution is 6.42 .



Hope this helped you...