Question

application questions: answer the following questions. show all work for calculations and include units and the correct number of significant figures. 1. what is the concentration (molarity) of a naoh solution if it reacts with 0.4578 g of khc₈h₄o₄ (khp) and the equivalence point is reached after 11.24 ml net volume of naoh solution has been added? the reaction is given by khc₈h₄o₄(s) + naoh(aq) → knac₈h₄o₄(aq) + h₂o(l) 2. below is a list of known acids for part b. you will be assigned to analyze an unknown which will be one of the acids listed. using the formula of the known acids, determine if the known acid is monoprotic, diprotic, or triprotic. calculate molar masses of the possible known acids by adding the atomic masses of each element in the acid using the periodic table. show the calculations for each molar mass in your lab notebook. known acid determine if known acid is monoprotic, diprotic, or triprotic molar mass, mₘ, for known acid (g/mol) citric acid (h₃c₆h₅o₇) sulfamic acid (hso₃nh₂) succinic acid (h₂c₄h₄o₄) tartaric acid (h₂c₄h₄o₆)

Explanation:

Step1: Calculate moles of KHP

The molar mass of $KHC_8H_4O_4$ (KHP):
$K$: 39.0983 g/mol, $H$: 1.0079 g/mol, $C$: 12.0107 g/mol, $O$: 15.9994 g/mol
$M_{KHP}=39.0983 + 1.0079+8\times12.0107 + 4\times1.0079+4\times15.9994$
$M_{KHP}=39.0983+1.0079 + 96.0856+4.0316+63.9976=204.221$ g/mol
The number of moles of KHP, $n_{KHP}=\frac{m}{M}=\frac{0.4578\ g}{204.221\ g/mol}=0.0022426\ mol$

Step2: Determine moles of NaOH

From the balanced chemical equation $KHC_8H_4O_4(aq)+NaOH(aq)
ightarrow KNaC_8H_4O_4(aq) + H_2O(l)$, the mole - ratio of KHP to NaOH is 1:1. So, $n_{NaOH}=n_{KHP}=0.0022426\ mol$

Step3: Calculate molarity of NaOH

The volume of NaOH, $V = 11.24\ mL=0.01124\ L$
The molarity of NaOH, $M=\frac{n}{V}=\frac{0.0022426\ mol}{0.01124\ L}=0.1995\ M\approx0.200\ M$

for part 2:

Citric acid ($H_3C_6H_5O_7$)

• Proticity: It has 3 acidic hydrogens, so it is triprotic.
• Molar mass:

$M=(3\times1.0079)+(6\times12.0107)+(5\times1.0079)+(7\times15.9994)$
$M = 3.0237+72.0642 + 5.0395+111.9958=192.1232\ g/mol$

Sulfamic acid ($HSO_3NH_2$)

• Proticity: It has 1 acidic hydrogen, so it is monoprotic.
• Molar mass:

$M=(1.0079)+(32.065)+(3\times15.9994)+(14.0067)+(2\times1.0079)$
$M=1.0079+32.065+47.9982+14.0067+2.0158 = 97.0936\ g/mol$

Succinic acid ($H_2C_4H_4O_4$)

• Proticity: It has 2 acidic hydrogens, so it is diprotic.
• Molar mass:

$M=(2\times1.0079)+(4\times12.0107)+(4\times1.0079)+(4\times15.9994)$
$M = 2.0158+48.0428+4.0316+63.9976=118.0878\ g/mol$

Tartaric acid ($H_2C_4H_4O_6$)

• Proticity: It has 2 acidic hydrogens, so it is diprotic.
• Molar mass:

$M=(2\times1.0079)+(4\times12.0107)+(4\times1.0079)+(6\times15.9994)$
$M=2.0158 + 48.0428+4.0316+95.9964=150.0866\ g/mol$

Answer:

The concentration of the NaOH solution is $0.200\ M$