Sodium sulfate salt weathering of porous building materials studied by NMR

Abstract

Sodium sulfate is known as one of the most destructive salts causing weathering. Many experiments on accelerated weathering tests show that major deterioration effects by weathering are caused by drying and wetting cycles of porous materials saturated with salt solution. In this study we have performed accelerated weathering tests with sodium sulfate in common building materials (fired-clay brick, Indiana limestone and Cordova limestone) measuring the concentration in the materials simultaneously with their expansion. The concentration of sodium sulfate solution is measured non-destructively using Nuclear Magnetic Resonance, while the expansion of the sample caused by crystal growth is measured with a fiber optic displacement sensor. The simultaneous measurement of solution concentration within a material and expansion allow assessment of crystallization pathways most responsible for damage during weathering, i.e., cycles of wetting and drying. It was shown by direct observation, that with rewetting of the partially dried samples, the present thenardite experiences a rapid partial transformation to decahydrate. Simultaneously with this transformation a rapid expansion of the sample was measured in situ.

Appendix: Calculation of salt mixture content during drying

Let us estimate the pore filling coefficient, i.e., the relative volume occupied by salts (thenardite and/or decahydrate) with respect to the pore volume (volume of the sample times porosity), $V_{0}$, initially filled with a solution. The pore filling coefficient of a particular salt varies from 0—no filling, to 1—complete filling and can be presented in the form:

$$\phi_{\text{t}} = \frac{{V^{\text{t}} }}{{V_{0} }};$$

(3)

$$\phi_{\text{d}} = \frac{{V^{\text{d}} }}{{V_{0} }};$$

(4)

where $\,\phi_{\rm t}$ and $\,\,\phi_{\rm d}$ are the pore filling coefficients, $\,V^{\rm t}$ and $\,V^{\rm d}$ are the crystal volumes of thenardite and decahydrate respectively. Then, the crystal volume of thenardite formed is $\,V^{\rm t} = V_{\rm m}^{\rm t} \,\Delta n$, where $\,V_{\rm m}^{\rm t} \, = 53$ cm³/mol is the molar volume of thenardite and $\,\Delta n$ is the amount (in moles) of thenardite precipitated by drying. If by rewetting all thenardite transforms to decahydrate, then the volume of decahydrate is $\,V^{\rm d} = V_{\rm m}^{\rm d} \,\Delta n$, where $\,V_{\rm m}^{\rm d} \, = 220$ cm³/mol.

The amount of precipitated thenardite by drying can be found from the difference between initial,$\,n_{0}$, and actual, $\,n$, amount of sodium sulfate in a solution:

$$\,\Delta n = n_{0} - n = \frac{{m_{0}^{\rm salt} - m^{\rm salt} }}{{M_{\rm salt} }};$$

(5)

where $m_{0}^{\rm salt}$ and $m^{\rm salt}$ are the initial and actual masses of thenardite in a solution. The mass of the solution,$\,m^{\rm sol}$, can be presented as a sum of the mass of water, $m^{\rm w}$, and of the sodium sulfate dissolved in a solution, $m^{\rm salt}$:

$$\,m^{\rm sol} = m^{\rm w} + m^{\rm salt} = n^{\rm w} M_{\rm w} + n^{\rm salt} M_{\rm salt} ;$$

(6)

where $n^{\rm w}$ and $n^{\rm salt}$ are the actual amounts (in moles) of water and salt in the solution respectively, $M_{\rm w}$ and $M_{\rm salt}$ are the molar masses of water and salt (in our case respectively 18 g/mol and 142.04 g/mol—thenardite). The concentration of a solution during the drying is:

$$c = \frac{{n^{\rm salt} }}{{m^{\rm w} }} = \frac{{n^{\rm salt} }}{{\theta \,m_{0}^{\rm w} }};$$

(7)

where $\theta$ is the moisture content and $m_{0}^{\rm w}$ is the initial mass of water in a solution. In the case, $\theta = 1$, when $n^{\rm salt}$ is equal to initial amount of salt in the solution (i.e., $n^{\rm salt} = n_{0}^{\rm salt}$), from Eq. (5) one will obtain the initial concentration of the solution, $c_{0}$. Therefore, the initial mass of water can be found from Eqs. (6) and (7) by the expression:

$$\,m_{0}^{\rm w} = \frac{{m_{0}^{\rm sol} }}{{\left( {1 + c_{0} \,M_{\rm salt} } \right)}};$$

(8)

where the mass of initial solution, $\,m_{0}^{\rm sol} = V_{0} \rho_{0}^{\rm sol}$, and the density of the initial sodium sulfate solution of 3 m is $\rho_{0}^{\rm sol} = 1.34$ g/cm³, and inner pore volume is a sample volume times porosity, hence $\,V_{0} = \rho_{\rm pore} \,V_{\rm sample}$, where $\,V_{\rm sample} = 3.14$ cm³.

Combining Eqs. (5)–(8), we can find the amount of thenardite that precipitated during the drying:

$$\,\Delta n = \rho_{\rm pore} \,V_{\rm sample} {\kern 1pt} \rho^{\rm sol} \,\,\left( {\frac{{c_{0} - c\,\theta }}{{1 + c_{0} \,M_{\rm salt} }}} \right);$$

(9)

where $c$ is the actual concentration of a solution and is measured by an NMR and presented in Fig. 4.

In the case, where not all thenardite is transformed to decahydrate, the concentration of a solution is expected to be between of the solubility of thenardite and decahydrate. Then the pore filling ratio of actually formed decahydrate (under assumption of homogenous distribution of a salt in the sample, i.e., Pe ≪ 1) can be in first order estimated by the relation of actual concentration of a solution and solubility concentrations of thenardite and decahydrate:

$$\,\,\phi_{{\rm d}}^{\rm act} \approx \left( {1 - \frac{{c - c_{\rm s}^{\rm d} }}{{c_{\rm s}^{\rm t} - c_{\rm s}^{\rm d} }}} \right)\,\phi_{\rm d} ;$$

(10)

In the case of all thenardite transformed to decahydrate, we return to initial expression (3).

The calculated initial pore volume,$\,V_{0}$, and pore filling coefficients $\,\phi_{\rm t}$ and $\,\,\phi_{{\rm d}}^{\rm act}$ are given in Table 1.