Although we don't know the exact pH because we don't have any concentrations we know that NaHCO3 is a weak acid, and NaOH is a strong base.
Thus we should get a solution that becomes more basic so the pH > 7.
that is due to acetic acid is a weak acid and sodium hydroxide is a strong base. hence when they are titrated pH increases
HCl + NaOH -----> NaCl + H2O
0.289 Moles is the molarity of an NaOH solution if 4.37 ml is titrated by 11.1 ml of 0.0904m hno3.
NaHCO3 (Sodium bicarbonate) [at room temperature] will produce a pH of around 8.
phenolphthalein.
that is due to acetic acid is a weak acid and sodium hydroxide is a strong base. hence when they are titrated pH increases
HCl + NaOH -----> NaCl + H2O
0.289 Moles is the molarity of an NaOH solution if 4.37 ml is titrated by 11.1 ml of 0.0904m hno3.
NaHCO3 (Sodium bicarbonate) [at room temperature] will produce a pH of around 8.
phenolphthalein.
0.050M x .05L = .0025 mol NaHCO3 0.10M x .0107L = .00107 mol NaOH Excess NaHCO3 = .0025-.00107 = 0.00143 pH = pKa2 + log(.00107/.00143) pH = 10.20
You need to know the volume of the weak acid being titrated so you can find how many moles of base are needed to match that of the acid.
8.78
H3NO3S + NaOH → Na3NO3S + H2O
You dont - adding NaOH increases pH.
1 millimolar = 0.001 M NaOH ( a base, remember ) - log(0.001 M NaOH) = 3 14 - 3 = 11 pH ----------
Sodium bicarbonate, NaHCO3 is alkaline so the pH will be greater than 7. However, the actual pH will depend on the concentration of the NaHCO3 solution.