6.0221367 rounded from the last digit would be 6.022137 but when using it calculations, you should just use 6.022 x 10^23.
Answer:
Use a ratio of 0.44 mol lactate to 1 mol of lactic acid
Explanation:
John could prepare a lactate buffer.
He can use the Henderson-Hasselbalch equation to find the acid/base ratio for the buffer.
![\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\3.5 = 3.86 + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\\log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 3.5 - 3.86 = -0.36\\\\\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 10^{-0.36} = \mathbf{0.44}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%20%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C3.5%20%3D%203.86%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%3D%203.5%20-%203.86%20%3D%20-0.36%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%3D%2010%5E%7B-0.36%7D%20%3D%20%5Cmathbf%7B0.44%7D)
He should use a ratio of 0.44 mol lactate to 1 mol of lactic acid.
For example, he could mix equal volumes of 0.044 mol·L⁻¹ lactate and 0.1 mol·L⁻¹ lactic acid.
0,35 kmol/m³ = 0,35 mol/dm³ = 0,35 mol/L
175 mL = 0,175 L
*-*-*-*-*-*-*-*-*-*-*-*
C = n/V
n = 0,35×0,175
n = 0,06125 mol
mCa(NO₃)₂: 40+(14×2)+(16×6) = 164 g/mol
1 mol --------- 164g
0,06125 ---- X
X = 10,045g
To prepare 175 mL of 0,35M solution, add 10,045g of calcium nitrate and add water to a volume of 175ml.