In standardization, 8.77 mL of $N a O H$ $NaOH$ neutralized 1.522 g of $K H P$ $KHP$ . Given the molar mass of $K H P$ $KHP$ is 204.22 g/mol, what is the molarity of the $N a O H$ $NaOH$ solution?

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In standardization, 8.77 mL of $N a O H$ $NaOH$ neutralized 1.522 g of $K H P$ $KHP$ . Given the molar mass of $K H P$ $KHP$ is 204.22 g/mol, what is the molarity of the $N a O H$ $NaOH$ solution?

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Answer 1 · 2017-06-13T03:14:38

$Molarity = 0.8500 \cdot m o l \cdot L^{- 1}$ $"Molarity"=0.8500*mol*L^-1$

Explanation:

$K H P$ $KHP$ , $potassium hydrogen phthalate$ $"potassium hydrogen phthalate"$ , $1, 2 - C_{6} H_{4} C O_{2} H C O_{2}^{-} K^{+}$ $1,2-C_6H_4CO_2HCO_2^(-)K^(+)$ , is the potassium salt of the diacid, $pthalic acid$ $"pthalic acid"$ , $1, 2 - C_{6} H_{4} {(C O_{2} H)}_{2}$ $1,2-C_6H_4(CO_2H)_2$ .

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It can be titrated as a base, or here as an acid. It is air stable, non-hygroscopic, and thus useful as a primary standard.

$1, 2 - C_{6} H_{4} C O_{2} H C O_{2}^{-} K^{+} + N a^{+}^{-} O H \to 1,2- C_{6} H_{4} C O_{2}^{-} N a^{+} C O_{2}^{-} K^{+} + H_{2} O$ $1,2-C_6H_4CO_2HCO_2^(-)K^(+)+Na^(+)""^(-)OH rarr "1,2-"C_6H_4CO_2^(-)Na^(+)CO_2^(-)K^(+)+H_2O$

And thus moles of $N a O H$ $NaOH$ $\equiv$ $-=$ $KHP$ $"KHP"$ , and thus............

$[N a O H] = \frac{\frac{1.522 \cdot g}{204.22 \cdot g \cdot m o l^{- 1}}}{8.77 \times 10^{3} \cdot L} = 0.8500 \cdot m o l \cdot L^{- 1}$ $[NaOH]=((1.522*g)/(204.22*g*mol^-1))/(8.77xx10^3*L)=0.8500*mol*L^-1$ .

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In standardization, 8.77 mL of $N a O H$ $NaOH$ neutralized 1.522 g of $K H P$ $KHP$ . Given the molar mass of $K H P$ $KHP$ is 204.22 g/mol, what is the molarity of the $N a O H$ $NaOH$ solution?

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In standardization, 8.77 mL of NaOHNaOHNaOH neutralized 1.522 g of KHPKHPKHP. Given the molar mass of KHPKHPKHP is 204.22 g/mol, what is the molarity of the NaOHNaOHNaOH solution?

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In standardization, 8.77 mL of $N a O H$ $NaOH$ neutralized 1.522 g of $K H P$ $KHP$ . Given the molar mass of $K H P$ $KHP$ is 204.22 g/mol, what is the molarity of the $N a O H$ $NaOH$ solution?